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presented  to  the 
UNIVERSITY  LIBRARY 
UNIVERSITY  OF  CALIFORNIA 
SAN  DIEGO 

by 

Dr.  George  F.  McEwen 


AN  ELEMENTARY  TREATISE  ON 


THEORETICAL  MECHANICS 


BY 


J.  H.  JEANS,  M.A.,  F.RS. 

Fellow  of  Trixity  College,  Cambridge  (England) 

AND  Professor  of  Applied  Mathematics 

IN  Princeton  University 


GIFT  OF 
mt  «GDRGE  F.  McQJfEWP 


GINN  &  COMPANY 

BOSTON  .  NEW  YORK  •  CHICAGO  •  LONDON 


Entered  at  Stationers'  Hall 


Copyright,  1907,  isv 
J.  H.  JEANS 


ALL  RIGHTS   RESERVED 

6G.12 


CINN  &  COMPANY  ■    PRO- 
PRIETORS •  BOSTON  •  U.S.A. 


PREFACE 

The  primary  aim  of  the  present  book  is  to  supply  for  students 
beginning  the  study  of  Theoretical  Mechanics  a  course  of  such  a 
nature  as  shall  emphasize  the  fimdamental  physical  principles  of 
the  subject.  Different  students  will  of  course  approach  the  study 
of  mechanics  with  different  mterests,  different  aims,  and  different 
amounts  of  mathematical  equipment,  so  that  it  may  not  be  possiljle 
to  produce  a  single  book  which  shall  exactly  fit  the  requirements 
of  every  class  of  student.  But  I  believe  that  all  students  of  me- 
chanics, no  matter  what  their  aims  and  intentions  may  be,  will 
be  in  the  same  position  m  one  respect,  namely  that  they  will  best 
begin  the  study  of  the  subject  by  trymg  to  acquire  a  firm  grasp 
of  the  physical  principles,  leaving  aside  at  first  all  mathematical 
developments  and  all  practical  applications,  except  m  so  far  as  these 
contribute  to  the  elucidation  of  the  fundamental  physical  prmciples. 

I  am  aware  that  this  belief  is  not  held  by  all  teachers  of 
mechanics,  some  of  whom  regard  the  laws  of  mechanics  simjily  as 
working  rules  to  be  acquired  as  rapidly  as  possible  for  their  utili- 
tarian value,  while  others  appear  to  regard  them  in  the  same  light 
as  the  rules  of  a  game,  the  game  consistmg  in  the  solution  of 
mathematical  puzzles,  most  of  wliich  have  no  conceivable  refer- 
ence to  the  facts  of  nature.  I  find  it  hard  to  believe  that  there 
can  be  any  considerable  class  of  students  for  whom  either  of  these 
points  of  view  is  the  best.  As  regards  the  former,  I  feel  that  a 
student  who  cannot  get,  or  does  not  wdsh  to  get,  a  clear  under- 
standing of  mechanical  prmciples  would  be  well  advised  not  to 
enter  a  profession  in  which  his  work  will  consist  in  the  handling 
of  mechanical  problems ;  and  as  regards  the  latter,  that  a  student 
who  wishes  merely  to  obtain  material  for  puzzle  solving  would  do 
better  to  turn  his  attention  to  chess  or  double  acrostics. 


iv  PKEFACE 

If  I  have  taken  some  space  to  express  my  private  convictions,  it 
is  because  the  method  I  have  embodied  in  the  present  book  arises 
directly  out  of  these  convictions.  Mathematical  analysis  is,  of 
course,  not  excluded  from  the  book,  because  without  mathematics 
there  can  be  no  serious  study  of  mechanics,  but  I  have  tried  to 
reduce  the  amount  of  mathematics  to  a  minimum,  and  1  have 
regarded  it  (in  the  present  book)  as  the  servant  and  not  as  the 
master.  Again,  practical  applications  of  mechanics  have  not  been 
excluded,  —  on  the  contrary,  these  have  been  introduced  wherever 
possible  as  illustrations  of  principles  or  results,  —  but  I  have  tried 
to  place  principles  first  and  applications  second.  And  problems 
have  not  been  excluded :  I  have  inserted  a  great  inimber,  because 
the  solution  of  problems  seems  to  me  to  be  the  one  and  indis- 
pensable way  of  ernphasizing  a  group  of  abstract  principles  and  of 
fixing  them  in  the  mmd  of  the  student.  But  I  have  regarded  the 
problems  as  an  adjunct  to  the  study  of  the  principles,  and  not  the 
principles  as  a  framework  round  which  to  build  problems. 

Besides  explaining  the  method  and  objects  of  a  book,  a  preface 
may  be  expected  to  explain  where  the  book  starts  and  where  it 
ends.  The  present  book  is  intended  to  start  from  the  very  begin- 
ning of  its  subject,  assuming  no  previous  knowledge  of  mechanics 
on  the  part  of  the  student.  The  question  of  how  much  knowledge 
of  mathematics  ought  to  be  assumed  has  been  a  more  difficult  one 
to  settle.  1  finally  decided  to  rely  as  little  as  possible  on  the  stu- 
dent's knowledge  of  trigonometry,  and  to  employ  the  calculus  as 
little  as  possible  in  the  earlier  chapters,  but  felt  that  the  subjects 
of  the  later  cliapters  could  not  be  advantageously  treated  without 
a  very  considerable  use  of  the  calculus.  Until  the  later  chapters 
the  use  of  the  calcidus  is  confined  almost  exclusively  to  unimpor- 
tant branches  and  extensions  of  the  subject,  and  to  the  working  of 
illustrative  examples.  Thus  a  student  who  has  no  knowledge  at 
all  of  the  calculus  will,  I  hope,  be  able  to  omit  the  sections  of  the 
book  in  which  it  is  used,  while  at  the  same  time  acquiring  a  con- 
siderable and  continuous  knowledge  of  the  essentials  of  theoretical 
mechanics. 


PREFACE  V 

The  point  at  which  the  book  ought  to  close  seemed  in  the  pres- 
ent instance  to  be  determined  by  the  method  of  the  book  itself. 
If,  as  I  beUeve,  a  study  of  physical  principles  ought  to  be  the  com- 
mon preliminary  to  the  study  of  every  branch  and  every  appli- 
cation of  mechanical  science,  then  the  book  might  clearly  try  to 
cover  all  tliis  common  ground,  and  ought  to  stop  at  the  point  at 
which  detailed  specialisation  becomes  feasible  and  profitable.  It 
ought,  in  fact,  to  cover  the  range  which  will  be  covered  by  all  stu- 
dents, and  stop  short  of  subjects  which  will  be  of  interest  or  impor- 
tance only  to  a  few.  Judged  by  this  criterion  the  book  will  perhaps 
be  thought  by  some  to  be  open  to  the  criticism  of  covering  too  much 
ground ;  it  may  be  thought  that  the  final  chapter  on  generalized 
coordinates  can  hardly  be  regarded  as  essential  to  the  student 
whose  study  of  mechanics  is  a  preliminary  to  his  entering  the  pro- 
fession of,  say,  engineering.  I  am  nevertheless  convinced  that,  even 
if  the  study  of  generalized  coordinates  is  not  absolutely  indispen- 
sable to  such  students,  it  is  of  extreme  value  and  ought  not  to  be 
neglected  by  a  student,  possessed  of  the  requisite  ability,  who  can 
possibly  find  time  for  it.  The  student  who  omits  it  shuts  himself 
off  from  a  point  of  view  which  sums  up  and  illuminates  the  whole 
of  dynamical  theory ;  at  the  same  time  he  denies  himself  the 
opportunity  of  studying,  or  at  least  of  fully  understanding,  the 
theory  of  electricity*  and  magnetism.  And  as  regards  the  student 
who  intends  to  continue  his  studies  in  the  direction  of  theoretical 
physics,  the  theory  of  generalized  coordmates  forms  so  essential 
a  preliminary  to  the  study  of  most  branches  of  physics  that  the 
advantages  of  including  a  short  treatment  of  this  subject  in  the 
preliminary  mechanics  course  will  hardly  be  disputed. 

Pkixcktox  J-  H.  JEAXS 

November,  1906 


CONTENTS 

Page 
Chapter  I.    Rest  and  Motion 1 

Introduction.  Motion  of  a  point.  Velocity.  Acceleration.  Vec- 
tors. 

Chapter  II.    Force  and  the  Laws  of  Motion 26 

Newton'.s  laws.  Frame  of  reference.  Laws  applicable  only  to  the 
motion  of  a  particle. 

Chapter  III.    Forces  acting  on  a  Single  Particle  ....      37 

Composition  and  resolution  of  forces.  Particle  in  equilibrium. 
Types  of  forces,  —  weight  of  a  particle,  tension  of  a  string,  reac- 
tion between  two  bodies.     Friction. 

Chapter  IY.    Statics  of  Systems  of  Particles 59 

Moments.  System  of  particles  in  equilibrium.  Forces  in  one  plane. 
Strings,  —  the  suspension  bridge,  the  catenary. 

Chapter  V.     Statics  of  Rigid  Bodies 90 

Rigidity.  Conditions  of  equilibrium  for  a  rigid  bodj^  Transmis.si- 
bility  of  force.  Composition  of  forces  acting  in  a  plane.  Parallel 
forces.    Couples.    Forces  in  space. 

Chapter  VI.     Center  of  Gravity 117 

Center  of  gravity  of  a  lamina.  Center  of  gravity  obtained  by  inte- 
gration.   Center  of  gravity  of  areas  and  volumes. 

Chapter  VII.    Work 145 

Measurement  and  units.  Work  done  against  a  variable  force. 
Work  done  in  .stretching  an  elastic  string.  Work  represented  by  an 
area.  The  principle  of  virtual  work.  Potential  energy.  Kinetic 
energy.    Conservation  of  energy.    Stable  and  unstable  equilibrium. 

vii 


viii  CONTENTS 

Page 
Chapter  VIII.    ^Iotiox  of  a  Particle  under  Constant  Forces    188 

Body  falling  under  gravity.  Motion  on  an  inclined  plane.  At- 
wood's  machine.  Motion  referred  to  a  moving  frame  of  reference. 
Frictional  reactions  between  moving  bodies.    Flight  of  projectiles. 

Chapter  IX.    Motion  of  Systems  of  Particles 220 

Equations  of  motion.  Conservation  of  momentum.  Motion  of 
center  of  gravity.    Kinetic  energy.    Impulsive  forces.    Elasticity. 

Chapter  X.    Motion  of  a  Particle  under  a  Variable  Force    254 

Equations  of  motion.  The  simple  pendulum.  Simple  hannonic 
motion.  The  cycloidal  pendulum.  Motion  of  a  particle  about  a 
center  of  force,  —  force  proportional  to  the  distance.  General  theory 
of  motion  about  a  center  of  force.    The  law  of  the  Inverse  square. 

Chapter  XI.     Motion  of  Rigid  Bodies 286 

Angular  velocity.  Kinetic  energy.  Radii  of  gyration.  Moment  of 
momentum.  General  theory  of  moments  of  inertia.  General  equa- 
tions of  motion  of  a  rigid  body.  Euler's  equations.  Rotation  of  a 
planet.    Motion  of  a  top. 

Chapter  XII.    Generalized  Coordinates 320 

Hamilton's  principle.  Principle  of  least  action.  Lagrange's  equa- 
tions. Small  o.scillations.  Stability  and  instability  of  equilibrium. 
Forced  oscillations.    The  canonical  equations. 

INDEX 361 


THEORETICAL   MECHAIflCS 


CHAPTER  I 
REST  AND  MOTION 

Introduction 

1.  Uniformity  of  nature.  If  we  place  a  stone  in  water,  it  will 
sink  to  the  bottom ;  if  we  place  a  cork  in  water,  it  will  rise  to 
the  top.  These  two  statements  will  be  admitted  to  be  true  not 
only  of  stones  and  corks  which  have  been  seen  to  sink  or  rise  in 
water  but  of  all  stones  and  corks.  Given  a  piece  of  stone  which 
has  never  been  placed  in  water,  we  feel  confident  that  if  we  place 
it  in  water  it  will  sink.  Wliat  justification  have  we  for  supposing 
that  tliis  new  and  untried  piece  of  stone  will  sink  in  water  ?  We 
know  that  millions  of  pieces  of  stone  have  at  different  times  been 
placed  in  water ;  we  know  that  not  a  single  one  of  these  has  ever 
been  known  to  do  anything  but  sink.  From  tliis  we  mfer  that 
nature  treats  all  pieces  of  stone  alike  when  they  are  placed  in 
water,  and  so  feel  confident  that  a  new  and  untried  piece  of  stone 
will  be  treated  by  the  forces  of  nature  in  the  same  way  as  the 
innumerable  pieces  of  stone  of  which  the  behavior  has  been 
tested,  and  hence  that  it  will  sink  in  water.  This  principle  is 
known  as  that  of  the  uniformity  of  nature  ;  what  the  forces  of 
nature  have  been  found  to  do  once,  they  will,  under  similar  condi- 
tions, do  again. 

2.  Laws  of  nature.  The  principle  just  stated  amounts  to  say- 
ing that  the  action  of  the  forces  of  nature  is  governed  by  certain 
laws ;  these  we  speak  of  as  laws  of  nature.  For  instance,  if  it 
has  been  found  that  every  stone  which  has  ever  been  placed  in 

1 


2  REST  a:n^d  motion 

water  has  sunk  to  the  bottom,  then,  as  has  already  been  said,  the 
principle  of  uniformity  of  nature  leads  us  to  suppose  that  every 
stone  which  at  any  future  time  is  placed  in  water  will  sink  to  the 
bottom ;  and  we  can  then  announce,  as  a  law  of  nature,  that  any 
stone,  placed  in  water,  will  sink  to  the  bottom. 

That  part  of  science  which  deals  with  the  laws  of  nature  is 
called  natural  science.  Natural  science  is  divided  into  two  parts, 
experimental  and  theoretical.  Experimental  science  tries  to  dis- 
cover laws  of  nature  by  observmg  the  action  of  the  forces  of 
nature  time  after  time. .  Theoretical  science  takes  as  its  material 
the  laws  of  nature  discovered  by  experimental  science,  and  aims  at 
reducing  them,  if  possible,  to  simpler  forms,  and  then  discovering 
how  to  predict  from  these  laws  what  the  action  of  the  forces  of 
nature  will  be  in  cases  wliich  have  not  actually  been  subjected  to 
the  test  of  experiment.  For  example,  experimental  science  dis- 
covers that  a  stone  sinks,  that  a  cork  floats,  and  a  number  of  sim- 
ilar laws.  From  these  theoretical  physics  arrives  at  the  simple  laws 
of  nature  which  govern  all  phenomena  of  sinking  or  floating,  and, 
going  further,  shows  how  these  laws  enable  us  to  predict,  before 
the  experiment  has  been  actually  tried,  whether  a  given  body  w^ill 
sink  or  float.  For  instance,  experimental  science  cannot  discover 
whether  a  50,000-ton  sliip  will  float  or  sink,  because  no  50,000- 
ton  ship  exists  with  which  to  experiment.  The  naval  architect, 
relying  on  the  uniformity  of  nature,  on  the  laws  of  nature  deter- 
mined by  experimental  science,  and  on  the  method  of  handling 
these  laws  taught  by  theoretical  science,  may  build  a  50,000-ton 
ship  with  every  confidence  that  it  will  behave  in  the  way  pre- 
dicted by  theoretical  science. 

3.  The  science  of  mechanics.  The  branch  of  science  known 
as  mechanics  deals  with  the  motion  of  bodies  in  space,  and  with 
the  forces  of  nature  which  cause  or  tend  to  cause  this  motion. 
The  laws  of  nature  which  govern  the  action  of  these  forces  and  the 
motion  of  bodies  have  long  been  known,  and  were  reduced  to  their 
•simplest  form  by  Newton.  Thus  we  may  say  that  experimental 
mechanics  is  a  completed  branch  of  science. 


MOTION  OF  A  POINT  3 

The  present  book  deals  with  theoretical  mechanics.  We  start 
from  the  laws  supplied  by  experimeutal  mechanics,  and  have  to 
discuss  how  these  laws  can  be  used  to  predict  the  motion  of  bodies, 
—  for  instance,  the  falling  of  bodies  to  the  ground,  the  firing  of 
projectiles,  the  motion  of  the  earth  and  the  planets  round  the  sun. 
An  important  class  of  problems  which  we  shall  have  to  discuss 
will  be  those  in  which  no  motion  takes  place,  the  forces  of  nature 
which  tend  to  cause  motion  being  so  evenly  balanced  that  no 
motion  occurs.    Such  problems  are  known  as  statical. 

Motion  of  a  Point 

4.  State  of  rest.  Before  we  can  reason  about  the  motion  of  a 
body  we  have  to  determine  what  is  meant  by  a  body  being  at 
rest.  In  ordinary  language  we  say  that  a  train  is  at  rest  when  the 
cars  are  not  moving  over  the  rails.  We  know,  however,  that  the 
train,  in  common  with  the  rest  of  the  earth,  is  not  actually  at  rest, 
but  moving  round  the  sun  with  a  great  velocity.  Again,  a  fly 
crawling  on  the  wall  of  a  railway  car  might  in  one  sense  be  said  to 
be  at  rest,  if  it  remained  standing  on  the  same  spot  of  the  wall. 
The  fly,  however,  would  not  actually  be  at  rest ;  it  would  share  in 
the  motion  of  the  train  over  the  country,  the  country  would  share 
in  the  motion  of  the  earth  round  the  sun,  and  the  sun  would 
share  in  the  motion  of  the  whole  solar  system  through  space. 

These  instances  will  show  the  necessity  of  attachmg  a  clear 
and  exact  meaning  to  the  conceptions  of  rest  and  motion.  Obvi- 
ouslj'  our  statements  would  have  been  exact  enough  if  we  had 
said  that  in  the  first  case  the  train  was  at  rest  relatively  to  the 
earth,  and  that  in  the  second  case  the  fly  was  at  rest  relatively  to 
the  car. 

5.  Frame  of  reference.  Thus  we  find  it  necessary,  before  dis- 
cussing rest  and  motion,  to  introduce  the  conception  of  a  frame  of 
reference.  The  earth  supplied  a  frame  of  reference  for  the  motion 
of  the  train,  and  when  a  train  is  not  moving  over  the  rails  we 
may  say  that  it  is  at  rest,  the  earth  being  taken  as  frame  of 


4  liEST  AND   MOTION 

reference.  So  also  we  could  say  that  the  fly  was  at  rest,  the  car 
being  taken  as  frame  of  reference.  Obviously  any  framework,  real 
or  imaginary,  or  any  material  body,  may  be  taken  as  a  frame  of  ref- 
erence, pro\dded  that  it  is  rigid,  i.e.  that  it  is  not  itself  changing 
its  shape  or  size. 

We  may  accordingly  say  that  a  point  is  at  rest  relatively  to  any 
frame  of  reference  when  the  distance  of  the  point  from  each  point 
of  the  frame  of  reference  remains  unaltered. 

6.  Motion  relative  to  frame  of  reference.  Having  specified  a 
frame  of  reference,  we  can  discuss  not  only  rest  but  also  motion 
relative  to  the  frame  of  reference.  When  the  train  has  moved 
a  mile  over  the  tracks  we  say  that  it  has  moved  a  mile  rela- 
tively to  its  frame  of  reference,  the  earth.  When  the  fly  has 
crawled  from  floor  to  ceilmg  of  the  car  we  say  that  it  has  moved, 
say,  eight  feet  relatively  to  its  frame  of  reference,  the  car. 

In  fixing  the  distance  traveled  by  the  fly  relatively  to  the  train  in  an 
interval  between  two  instants  t^,  t^,  we  notice  that  the  actual  point  from 
which  the  fly  started  is,  say,  a  mile  behind  the  present  position  of  the 
train  ;  but  the  point  from  which  we  measure  is  the  point  which  occupies 
the  same  position  in  the  car  at  time  <2  as  this  point  did  at  time  t^.  So,  in 
general,  to  fix  the  distance  moved  relatively  to  a  given  frame  of  reference 
in  the  interval  between  times  t^  and  t^,  we  first  find  the  point  A  which  stands 
in  the  same  position  relative  to  the  frame  of  reference  at  time  t^  as  did 
the  point  from  which  the  moving  point  started  at  time  t^.  The  distance 
from  this  point  A  to  the  point  B,  which  is  occupied  by  the  moving  point  at 
instant  ^2?  is  the  distance  moved  relatively  to  the  moving  frame  of  reference. 

By  the  motion  of  a  particle  B  relative  to  a  particle  J,  is  meant 
the  motion  of  B  relative  to  a  frame  of  reference  moving  with  A. 

7.  Composition  of  motions.  Suppose  that  in  a  given  time  the 
moving  point  moves  a  certain  distance  relatively  to  its  frame  of 
reference,  wliile  tliis  frame  of  reference  itself  moves  some  other 
distance  relatively  to  a  second  frame  of  reference,  —  as  will,  for 
instance,  occur  if  a  fly  climbs  up  the  side  of  a  car  wliile  the  car 
moves  relatively  to  the  earth. 

Let  us  suppose  that  there  is  a  frame  of  reference  mo\'ing  in 
the  plane  of  the  paper  on  which  fig.  1  is  drawn,  and  that  the 


MOTION  OF  A  POINT 


Fig.  1 


paper  itself  supplies  a  second  frame  of  reference.  Suppose  that 
the  moving  point  starts  at  A,  and  that  during  the  motion  that 
point  of  the  first  frame  of  reference  which  originally  coincided 
with,  the  moving  point  has  moved 
from  A  to  B,  while  the  point  itself 
has  moved  to  C.  Then  the  line  AB 
represents  the  motion  of  frame  1 
relative  to  frame  2,  while  BC  repre-  ''"^ 
sents  the  motion  of  the  moving  point 
relative  to  frame  1.  The  whole  mo- 
tion of  the  point  relative  to  frame  2  is  represented  by  AC.  The 
motion  .^C  is  said  to  be  compounded  of  the  two  motions  AB,  BC, 
or  is  said  to  be  the  resultant  of  the  two  motions.    Thus : 

If  a  point  moves  a  distance  BC  relatively  to  frame  1,  while 
frame  1  moves  a  distance  AB  relatively  to  frame  2,  the  resultant 
motion  of  the  poiiit  relative  to  frame  2  will  be  the  distance  AC, 
obtained  by  taking  the  tivo  distances  AB,  BC  and  placviuj  tliem  in 
p)Osition  in  such  a  way  that  the  point  B  at  which  the  one  ends  is 
also  the  point  at  which  the  other  begins. 

There  is  a  second  way  of  compounding  two  motions.  Let  x,  y 
represent  the  two  motions.  The  rule  already  obtained  directs  us  to 
construct  a  triangle  ABC,  to  have  x,  y  for  the  sides  AB,  BC,  and 
then  AC  will  be  the  motion  required.    Having  constructed  such  a 

triangle  ABC,  let  us 
complete  the  paral- 
lelogram ABCD  by 
drawing  AD,  CD 
parallel  to  the  side 
of  the  triangle. 
Then  AD,  being 
equal  to  BC,  will  also  represent  the  motion  y,  so  that  we  may  say 
that  the  two  edges  of  the  parallelogram  wliich  meet  in  A  represent 
the  two  motions  to  be  compounded,  while  the  diagonal  AC  through 
A  has  already  been  seen  to  represent  the  resultant  motion.  Thus 
we  have  the  following  rule  for  compoundmg  two  motions  x,  y : 


Fig. 


6  KE8T  AND  MOTION 

Construct  a  parallelogram  ABCD  such  that  the  two  sides  AB,  AD 
which  meet  in  A  represent  the  two  motions  x,  y  to  he  compounded, 
as  regards  both  magnitude  and  direction;  then  the  diagonal  AC 
which  passes  through  A  will  represent  the  resultant  obtained  by 
compounding  these  tivo  motions. 

Velocity 

8.  Uniform  and  variable  velocity.  Velocity  means  simply 
rate  of  motion.  It  may  be  either  uniform  or  variable.  If  a  point 
moves  in  such  a  way  that  a  feet  are  described  in  each  second  of 
its  motion,  no  matter  which  second  we  select,  we  say  that  the 
velocity  of  the  point  is  a  uniform  velocity  of  a  feet  per  second. 
If,  however,  the  point  moves  a  feet  in  one  second,  b  feet  in  another, 
c  feet  in  a  third,  and  so  on,  we  cannot  say  that  any  one  of  the 
quantities  a,  b,  or  c  measures  the  velocity.  The  velocity  is  now 
said  to  be  variable  :  it  is  different  at  different  stages  of  the  motion. 
To  define  the  velocity  at  any  instant,  we  take  an  infinitesimal  in- 
terval of  time,  clt  and  measure  the  distance  ds  described  in  tliis 

ds 
time.    We  then  define  the  ratio  —  to  be  the  velocity  at  the  instant 

dt  , 

CIS 

at  which  the  interval  clt  is  taken.    If  the  velocity  is  uniform,  — 

dt 

is  the  space  described  in  unit  time,  and  so  the  present  definition 

of  velocity  becomes  the  same  as  that  already  given. 

Average  velocity.     If  a  point  moves  with  variable  velocity,  and 

describes  a  distance  of  a  feet  in  t  seconds,  we  speak  of  -  as  tlie 

"  average  velocity "  of  the  moving  point  during  the  time  t.  This 
average  velocity  is  the  velocity  which  would  have  to  be  possessed 
by  an  imaginary  point  moving  with  uniform  velocity,  if  it  were  to 
cover  the  same  distance  in  time  t  as  the  actual  point  moving  with 
variable  velocity. 

Units.  In  measuring  a  velocity  we  need  to  speak  in  terms  of  a 
unit  of  length  and  of  a  unit  of  time ;  for  instance,  in  saying  that 
a  point  has  a  velocity  of  a  feet  per  second  we  have  selected  the  foot 


VELOCITY  7 

as  unit  of  length  and  the  second  as  unit  of  time.    We  can  find  the 
amount  of  tliis  same  velocity  in  other  units  by  a  simple  proportion. 

Thus  suppose  it  is  required  to  express  a  velocity  of  a  feet  per  second 
in  terms  of  miles  and  hours. 

The  point  moves  a  feet  in  one  second,  and  therefore  a  x  60  x  60  feet  in 
one  hour,  and  therefore 

a  X  60  X  60      1.5  a 


3  X  1760  22 


miles 


in  one  hour.    Thus  the  velocity  is  one  of  -— -  miles  per  hour. 


EXAMPLES 

1.  A  railway  train  travels  a  distance  of  918  miles  in  18  hours.  What  is  its 
average  velocity  in  feet  per  second  ? 

2.  Compare  the  velocities  of  a  train  and  an  automobile  which  move  uni- 
formly, the  former  covering  100  feet  a  second  and  the  latter  1.500  yards  a  minute. 

3.  A  man  runs  100  yards  in  9^  seconds.  What  is  his  average  speed  in  miles 
per  hour  ? 

4.  The  two  hands  of  a  town  clock  are  10  and  7  feet  long.  Find  the  velocities 
of  their  extremities. 

5.  Taking  the  diameter  of  the  earth  as  7927  miles,  what  is  the  velocity  in 
foot-second  units  of  a  man  standing  at  the  equator  (in  consequence  of  the  daily 
revolution  of  the  earth  about  its  axis)  ? 

6.  Two  trains  230  and  440  feet  long  respectively  pass  each  other  on  parallel 
tracks,  the  former  moving  with  twice  the  speed  of  the  latter.  A  passenger  in 
the  shorter  train  obsei'\'es  that  it  takes  the  longer  train  three  seconds  to  pass 
him.    Find  the  velocities  of  both  trains. 

9.  Composition  of  velocities.  All  motion,  as  we  have  seen,  must 
be  measured  relatively  to  a  frame  of  reference.  Thus  velocity, 
or  rate  of  motion,  must  also  be  measured  relatively  to  a  frame 
of  reference.  A  point  may  have  a  certam  velocity  relative  to  a 
frame  of  reference,  while  the  frame  of  reference  itself  has  another 
velocity  relative  to  a  second  frame.  It  may  be  necessary  to  find 
the  velocity  of  the  moving  pomt  with  reference  to  the  second 
frame,  in  other  words,  to  comjyound  the  two  velocities. 

To  do  this  we  consider  the  motions  wliich  take  place  during 
an  infinitesimal  interval  of  time  dt.  Let  the  moving  point  have 
a  velocity  ?;j  in  a  direction  AB  relative  to  the  first  frame,  while 


8 


REST  AXU  MOTION 


the  frame  has  a  velocity  v^  in  a  direction  AC  relative  to  the 
second  frame.  Then  in  time  dt  the  moving  point  describes  a  dis- 
tance i\dt,  say  the  distance  AD,  along  AB  relative  to  the  first 
frame,  wliile  the  frame  itself  describes  a  distance  v^dt,  say  AE, 
along  AC  relative  to  the  second  frame.    Let  AF  be  the  diagonal 

of  the  parallelogram  of  which  AB,  AE 
are  two  edges;  then  AF  will  be  the 
resultant  motion  of  the  point  in  time 
dt  relative  to  the  second  frame.  Since 
the  moving  point  describes  a  distance 
AF  hi  time  dt,  the  resultant  velocity 


Fig.  3 


AF 

will  be 

dt 


Let  us  now  agree  that  velocities  are 
to  be  represented  by  straight  lines,  the  direction  of  the  line  being 
parallel  to  that  of  the  velocity  and  its  length  being  proportional 
to  the  amount  of  the  velocity,  the  lengths  being  drawn  according 
to  any  scale  we  please ;  for  example,  we  might  agree  that  every 
inch  of  length  is  to  represent  a  velocity  of  one  foot  per  second,  in 
wliich  case  a  velocity  of  three  feet  a  second  will  be  represented  by  a 
line  three  inches  long  drawn  parallel  to  the  direction  of  motion. 

In  fig.  3  let  Ap,  Aq  represent  the  velocities  t\,  v^  drawn  on  any 
scale  we  please.    Since  the  scale  is  the  same  for  both,  we  have 

Ap  :  Aq  =  v^ :  v^. 

Now  AE  =  v^dt,  AD  =  i\dt,  so  that 

AE:AD  =  Vo_:  v^, 
and  hence  Ap  :  Aq  =  AE :  AD. 

If  we  complete  the  parallelogram  Aprq,  the  diagonal  Ar  will  pass 
through  F,  and  we  shall  have 

Ar  :  Ap  =  AF :  AE. 

If  V  is  the  resultant  velocity,  it  has  already  been  seen  that 

AF 

dt 


VELOCITY 


9 


so  that  AF :  AIJ  =  Vdt :  v^dt 

=  V:v„ 
and  hence  Ar  :  Ap  =  V :  v^. 

Thus  Ar  represents  the  magnitude  of  the  velocity  V  on  the 
same  scale  as  that  on  which  Ap  represents  the  velocity  v^.  Also 
since  Ar  is  in  the  direction  of  AF,  the  resultant  motion,  we  see 
that  Ar  represents  the  velocity  Fboth  in  magnitude  and  direction. 
We  have  accordingly  proved  the  following  theorem : 

Theorem.  If  ttvo  velocities  are  represented  in  magnitude  and 
direction  ty  the  two  sides  of  a  parallelograin  which  start  from  any 
point  A,  then  their  resultant  is  represented  in  magnitude  and  direc- 
tion on  the  same  scale  hy  the  diagonal  of  the  parallelogram  which 
starts  from  A. 

This  theorem  is  known  as  the  parallelogram  of  velocities.  We 
may  illustrate  its  meaning  by  two  simple  examples. 

1.  Suppose  that  a  carriage  is  moving  on  a  level  road  witli  velocity  V. 
As  a  first  frame  of  reference  let  lis  take  the  body  of  the  carriage;  as 
a  second  frame  take  the  road  itself.  The  velocity  of  frame  1  relative  to 
frame  2  is  then  V.  Relatively  to  frame  1,  the  center  of  any  wheel  P  is 
fixed,  so  that  any  point 

on  the  rim  describes  JI      ^  Q^ 

a  circle  about  P.  Rela- 
tively to  frame  1  the 
road  is  moving  backward 
with  velocity  V,  so  that 
if  there  is  to  be  no  slip- 
ping between  the  rim  and 
the  road,  the  velocity  of 
any  point  on  the  rim,  rel- 
ative to  the  first  frame 
(the  carriage),  must  be  V. 

Thus  the  velocity  of  any  yig  4 

point  Q  on  the  rim  rela- 
tive to  frame  1  will  be  a  velocity  V  along  the  tangent  (2T.  Representing 
this  by  the  line  QT,  the  velocity  of  the  carriage  relative  to  the  road  is 
represented  by  an  equal  line  QH  parallel  to  the  road.  Thus  the  resultant 
velocity  of  the  point  Q  is  represented  by  the  diagonal  QS  of  the  parallelo- 
gram QHST.    Clearly  its  direction  bisects  the  angle  HQT.    Let  L  be  the 


10  KEST  AND  MOTIOX 

lowest  point  of  the  wheel,  and  let  X  complete  the  parallelogram  QPLX. 
Obviously  this  parallelogram  is  similar  to  the  parallelogram  QTSH,  corre- 
sponding lines  in  tlie  two  parallelograms  being  at  right  angles.    Thus 

QS  :  QT  =  QL  :  QP. 

So  that  on  a  scale  in  which  the  velocitj'  of  the  carriage  is  rejiresented  in 
magnitude  by  QP,  the  radius  of  the  wheel,  the  velocity  of  the  point  Q  will 
be  represented  by  QL.  Thus  the  velocities  of  the  different  points  on  the 
rim  are  proportional  to  their  distances  from  L,  their  directions  being  in 
each  case  perpendicular  to  the  line  joining  the  point  to  L. 

2.  A  battle  ship  is  steaming  at  18  knots,  and  its  guns  can  fire  projectiles 
with  velocities  of  2000  feet  per  second  relative  to  the  ship.    How  must 
the  gims  be  pointed  to  hit  an  object  the  direction  of 
which  from  the  ship  is  perpendicular  to  that  of  the 
shijj's  motion? 

hetAB  be  the  direction  of  the  ship's  motion,  and 
let  us  suppose  the  gun  pointed  in  a  direction  AC. 
Then  the  velocity  of  the  shot  relative  to  the  ship 
can  b"e  represented  b}^  a  line  Ap  along  A  C,  while  that 
of  the  ship  relative  to  the  sea  can  be  represented  by 
a  line  Aq  along  AB.  Completing  the  parallelogram 
Aprq,  we  find  that  the  diagonal  Ar  will  represent  the 
velocity  of  the  shot  relative  to  the  sea  in  magnitude  and  direction.  Hence 
Ar  must,  from  the  data  of  the  question,  be  at  right  angles  to  AB.  li  0  is 
the  angle  pAr  through  which  the  gun  must  be  turned  after  sighting  the 
object  to  be  hit,  we  have 

pr  _  velocity  of  ship 


sm^ 


Ap      velocity  of  firing  of  shot 


The  velocity  of  the  ship  is  18  knots,  or  18  nautical  miles  per  hour. 
Now  1  nautical  mile  =  1.1515  ordinary  miles  =  6080  feet,  so  that  a 
velocity  of  18  knots  is  equal  to  109,440  feet  per  hour,  or  30.4  feet  per 

second.    Thus  sin  0  =  '- — ^  =  .0152,  whence  we  find  that  6  =  0°  52'  16". 
2000 


Triangle  of  Velocities 

10.  We  can  also  compound  velocities  by  a  rule  known  as  the 
triangle  of  velocities.  In  fig.  .3  the  two  velocities  were  represented 
by  Ap,  Aq,  and  their  resultant  by  Ar.  The  two  velocities,  how- 
ever, might  equally  well  have  been  represented  by  Ap,  pr,  and 
their  resultant  by  Ar,  from  wliich  we  obtain  the  following  rule:   ' 


Fig.  6 


VELOCITY  11 

If  two  velocities  are  represented  hy  the  two  sides  of  a  triangle 
taken  in  order,  their  resultant  will  he  represented  hy  the  third  side, 
taken  in  the  direction  from  the  first  side  to  the  second  side. 

For  example,  let  OP^,  OP^  be  two  lines  drawn  through  0  to 
represent,  on  any  scale,  the  velocities  of  a  mov- 
ing point  at  instants  t^,  t^.    Then  P^P^,  will,  on 
the  same  scale,  represent  the  additional  velocity- 
acquired  by  the  pomt  iii  this  interval. 

For  we  can  imagine  a  frame  mo\ing  with 
the  uniform  velocity  OP^  of  the  particle  at 
instant  t^.  The  velocity  OP^  at  instant  t^  may 
be  supposed  compounded  of  the  velocity  OP^ 
of  the  frame  and  a  velocity  P^Pz  relative 
to  the  frame.    Obviously  this  latter  is  the  increase  of  velocity. 

EXAMPLES 

1.  A  car  is  running  at  14  miles  an  liour,  and  a  man  jumps  from  it  with  a 
velocity  of  8  feet  per  second  in  a  direction  making  an  angle  of  30°  with  the 
direction  of  the  car's  motion.    What  is  his  velocity  relative  to  the  ground  ? 

2.  A  railway  train,  moving  at  the  rate  of  60  miles  an  hour,  is  struck  by  a 
bullet,  which  is  fired  horizontally  and  at  right  angles  to  the  train  with  a  velocity 
of  440  feet  a  second.  Find  the  magnitude  and  direction  of  the  velocity  with 
which  the  bullet  appears  to  meet  the  train  to  a  person  inside. 

3.  A  ship  whose  head  points  northeast  is  steaming  at  the  rate  of  12  knots  in 
a  current  which  flows  southeast  at  the  rate  of  6  knots.  How  far  will  the  ship 
have  gone  in  2\  hours  ? 

4.  A  train  is  traveling  at  the  rate  of  30  miles  an  hour,  and  rain  falls  with 
a  velocity  of  22  feet  per  second  at  an  angle  of  30°  with  the  vertical  in  the  same 
direction  as  the  motion  of  the  train.  Find  the  direction  of  the  spla.shes  made 
on  the  windows  by  the  raindrops. 

5.  A  steamer's  course  is  due  south,  and  its  speed  is  20  knots ;  the  wind  is 
from  the  west,  but  the  line  of  smoke  from  the  steamer  is  observed  to  point  in  a 
direction  30°  east  of  north.    What  is  the  velocity  of  the  wind  ? 

6.  A  man  rows  across  a  stream  a  mile  wide,  pointing  his  boat  upstream  at 
an  angle  of  30°  with  the  bank.  How  long  does  he  take  to  cross,  if  he  rows  with 
a  velocity  of  4  miles  an  hour  and  if  the  current  has  an  equal  velocity? 

7.  A  stream  has  a  current  velocity  a,  and  a  man  can  row  his  boat  with  a 
velocity  b.  In  what  direction  must  he  row,  if  he  is  to  land  at  a  point  exactly 
opposite  his  starting  point?  And  in  wliat  direction  must  he  row  so  as  to  cross 
in  the  shortest  time  ? 


12  REST  AND  MOTION 

8.  A  ship  whose  head  is  pointing  due  soutli  is  steaming  across  a  current  run- 
ning due  west ;  at  the  end  of  two  hours  it  is  found  tliat  the  ship  lias  gone  36  miles 
in  the  direction  15°  west  of  south.    Find  the  velocities  of  the  ship  and  current. 

9.  A  pei-son  traveling  eastward  at  the  rate  of  3  miles  an  hour  finds  that  the 
wind  seems  to  blow  directly  from  the  north  ;  on  doubling  his  speed  it  appears 
to  come  from  the  northeast.    Find  the  direction  of  the  wind  and  its  velocity. 


ACCELERxVTION 

11.  Acceleration  is  rate  of  increase  of  velocity.  If  we  find  that 
the  velocity  of  a  moving  point  increases  by  an  amount  /  in  a  sec- 
ond, no  matter  wliich  second  is  selected,  we  say  that  the  motion 
of  the  point  has  a  uniform  acceleration  /  per  second.  For  instance, 
a  stone  or  other  body  falling  under  gravity  is  foimd  to  increase 
its  velocity  by  a  certain  constant  velocity  /  per  second,  where  / 
denotes  a  velocity  of  about  32  feet  per  second.  Thus  we  say  that 
a  falling  stone  has  a  uniform  acceleration  of  /  per  second,  or  of 
about  32  feet  per  second  per  second. 

Generally,  however,  an  acceleration  will  not  be  uniform ;  the 
rate  of  increase  of  velocity  will  be  different  at  different  stages  of 
the  journey.  To  find  the  acceleration  at  any  instant,  we  observe 
the  change  in  velocity  during  an  infinitesimal  interval  dt  of  time. 

dv 
If  dv  is  the  increase  of  velocity,  we  say  that  —  is  the  acceleration 

at  the  instant  at  wliich  dt  is  taken.  An  acceleration  will  of  course 
have  sign  as  well  as  magnitude,  for  the  velocity  may  be  either 
increasing  or  decreasing.  AVlien  the  velocity  is  decreasing,  the 
acceleration  is  reckoned  with  a  negative  sign.  A  negative  accelera- 
tion is  spoken  of  as  a  retardation.  Thus  a  retardation  /  means 
that  the  velocity  is  diminished  by  an  amount  /  per  unit  of  time. 

EXAMPLES 

1.  A  workman  fell  from  the  top  of  a  building  and  struck  the  ground  in 
4  seconds.  With  what  velocity  did  he  strike  the  ground,  the  acceleration  due  to 
gravity  being  32  feet  per  second  per  second  ? 

2.  A  train  has  at  a  given  instant  a  velocity  of  30  miles  an  hour,  and  moves 
with  an  acceleration  of  1  foot  per  second  per  second.  Find  its  velocity  after 
20  seconds. 


ACCELEKATION 


3.  A  train  comes  to  rest  after  the  brakes  have  been  applied  for  ten  seconds. 
If  the  retardation  was  8  feet  per  second  per  second,  what  was  the  velocity  of 
the  train  when  tiie  brakes  were  first  drawn  ? 

4.  How  long  does  it  take  a  body  starting  with  a  velocity  of  22  feet  per  second 
and  moving  with  an  acceleration  of  G  feet  per  second  per  second,  to  acquire  a 
velocity  of  60  miles  an  hour  ? 

5.  Two  bodies  start  at  the  same  instant  with  velocities  u  and  v  respectively; 
the  motion  of  the  first  undergoes  a  retardation  of /feet  per  second  per  second, 
while  that  of  the  second  is  uniform.  How  far  will  the  second  have  gone  by  the 
time  that  the  first  comes  to  rest  ? 

6.  A  body  starting  from  rest  moves  for  4  seconds  with  a  uniform  accelera- 
tion of  8  feet  per  second  per  second.  If  the  acceleration  then  ceases,  how  far 
will  the  body  move  in  the  next  5  seconds  ? 

7.  A  train  has  its  speed  reduced  from  40  miles  an  hour  to  30  miles  an  hour 
in  5  seconds.  If  the  retardation  be  unif oi-m,  for  how  much  longer  will  it  travel 
before  coming  to  rest  ? 

8.  A  body  falling  under  gravity  has  an  acceleration  of  32.2  feet  per  second 
per  second.  Express  this  acceleration  when  the  units  are  (a)  centimeter, 
second  ;   (b)  mile,  hour. 

12.  Parallelogram  of  accelerations.  Theorem.  Let  the  velocity/ 
of  a  point  he  comjMunded  of  two  velocities  v^,  v^  along  given  direc- 
tions, and  let  these  velocities  be  variable,  their  accelerations  being 
fv  fz-  Then  if  tivo  lines  be  drawn  in  the  direction  of  the  velocities, 
to  represent  f^,  f^  on  any  scale,  the  resultant  acceleration  tvill  be 
represented  on  the  same  scale  by  the  diagonal  of  the  parallelogram 
of  which  these 
lines  are  edges. 

To  prove  the 
theorem,  we 
consider  the 
motion  dur- 
ing any  small 
interval  dt 
at  wliich  the 
component  ac- 
celerations are 

fi,  /j.  In  fig.  7  let  AB,  ^C  represent  the  two  velocities  v^,  v.^  at  the 
beginning  of  this  interval.  Let  BB',  CC  represent,  on  the  same 
scale,  the  infinitesimal  increments  in  velocity  in  the  interval  dt, 


Fig. 


14  REST  AND  MOTION 

nsimely  f^dt,f^dt.    Then  AB' ,  AC  will  represent  the  velocities  at 
the  end  of  the  interval  dt. 

In  the  figure  the  lines  BDF,  B'ED',  CDE,  C'FD'  are  drawn 
parallel  to  AB  and  A  C.  Thus  AD  represents  the  resultant  velocity 
at  the  beginning  of  the  interval  dt,  and  AD'  that  at  the  end  of  the 
interval.  The  velocity  AD'  can  be  regarded  as  compounded  of  the 
two  velocities  AD,  DD' ,  and,  as  in  §  1 0,  DD'  represents  the  incre- 
ment in  velocity  in  time  dt.  Thus,  if  F  is  the  resultant  acceleration, 
the  line  DD'  will  represent  a  velocity  Fdt.  On  the  same  scale  DE, 
DF  represent  velocities  f^dt,  f^dt,  and  DED'F  is  a  parallelogram. 

If  OF^,  OF^  (fig.  8)  represent  the  accelerations  f-^,/^  on  any  scale, 
and  if  OG  is  the  diagonal  of  the  completed  parallelogram,  we 
clearly  have  OF^ :  OF^  =  A  :/2  =  DE :  DF, 
so  that  the  parallelograms  OF^GF^  (in 
fig.  8)  and  DED'F  (in  fig.  7)  wdl  be  simi- 
lar and  similarly  situated.  Thus 
OG:OF,  =  DD' :  DE  =  Fdt  :f^dt  =  F:f„ 
so  that  OG  represents  the  acceleration  F 
on  the  same  scale  as  that  on  which  OF^, 
OF^  represent /i,/, ;  and  OG,  being  parallel 
to  DD',  will  also  represent  the  direction  of  F,  proving  the  theorem. 
Clearly  the  acceleration  at  any  instant  need  not  be  in  the  same 
direction  as  the  velocity.  In  fig.  7  the  directions  AD,  AD'  repre- 
sent velocities  at  the  beginnmg  and  end  of  the  interval  dt.  Wlien 
in  the  limit  we  take  dt  =  0,  these  lines  coincide,  and  the  direction 
of  the  velocity  at  the  instant  at  which  dt  is  taken  is  that  of  AD. 
The  direction  of  the  acceleration  at  this  instant  is,  however,  DD'. 

As  an  illustration  of  this,  let  us  consider  the  motion  of  a  particle  mov- 
ing uniformly  in  a  circle  ;  e.g.  a  point  on  the  rim  of  a  wheel,  turning  with 
uniform  velocity  V  about  its  center. 

Let  A,  B  (fig.  9)  be  the  positions  of  the  point  at  two  instants,  let  the 
tangents  at  A,  B  meet  in  C,  and  let  D  complete  the  parallelogram  A CBD. 

The  velocity  at  the  first  instant  is  a  velocity  V  along  A  C.  Let  us  agree 
to  represent  this  by  the  line  A  C  itself.  At  the  second  instant  the  velocity 
is  a  velocity  V  along  CB  ;  this  may,  on  the  same  scale,  be  represented  by 
the  line  CB,  or  more  conveniently  by  AD.    Since  AC,  AD  represent  the 


ACCELERATION  15 

velocities  at  the  two  instants,  the  line  CD  will  represent  the  change  in 
velocity  between  these  two  instants. 

Now  let  the  two  instants  differ  only  by  an  infinitesimal  interval  <lt,  so 
that  the  points  A ,  B  coincide  except  for  an  infinitesimal  arc  Vdl.  In  the 
figure,  CD  passes  through 
P  wherever  A ,  B  are  on  the 
circle,  so  that  when  B  is 
made  to  coincide  witli  A, 
CD  coincides  with  the  ra- 
dius through  A.  But  if  F 
is  the  acceleration  of  the 
moving  point,  the  chan;.';e 
in  velocity  prod  viced  in 
time  (It  must  be  Fdt.  Thus  CD  represents  the  change  of  velocity  Fdl  in 
direction  and  magnitude,  so  that  the  change  of  velocity,  and  hence  the 
acceleration  at  A ,  is  along  the  radius  at  A . 

Here,  then,  we  have  a  case  in  which  the  acceleration  is  at  right  angles 
to  the  velocity. 

To  find  the  magnitude  of  the  acceleration,  we  notice  that  CD  =  2  CE, 
and  that,  by  similar  triangles, 

EC  :  CB  =  BE  :  BP. 

Now  EC,  or  I  CD,  represents  the  velocity  i  Fdt,  while  CB  on  the  same 
scale  represents  the  velocity  V. 

Thus  i  Fdt  :  V  =  BE  :  BP. 

In  the  limit  when  BA  is  very  small,  BE,  or  \BA,  becomes  identical 
with  half  of  the  arc  BA  of  the  circle,  and  therefore  with  -J-  Vdt.  Thus,  if  a 
is  the  radius  of  the  circle, 

^Fdt  :  r=iVdt  -.a, 

giving  F  =  —  as  the  amount  of  the  acceleration. 

EXAMPLES 

1.  A  windmill  has  sails  20  feet  in  length,  and  turns  once  in  ten  seconds. 
Find  the  acceleration  of  a  point  at  the  end  of  a  sail. 

2.  A  wheel  of  radius  3  feet  spins  at  the  rate  of  10  revolutions  a  second  and 
is  at  the  same  time  falling  freely  with  an  acceleration  of  32  feet  per  second 
per  second  due  to  gravity.  Find  the  resultant  accelerations  of  the  different 
points  on  the  rim  of  the  wheel. 

3.  Taking  the  eartli  to  have  an  equatorial  diameter  of  7927  miles,  find  the 
acceleration  towards  the  earth's  center  of  (a)  a  point  at  rest,  relative  to  the 
earth's  surface,  on  the  equator;  (6)  a  body  falling  under  gravity  at  the  equator, 


16  REST  AND  MOTION 

■with  an  acceleration,  relative  to  the  earth's  surface,  of  32.09  feet  per  second 
per  second. 

4.  Supposing  that-the  moon  describes^  circle  of  radius  240,000  miles  round 
the  earth  in  29^  days,  find  its  acceleration  towards  the  earth. 

5.  Assuming  that  the  planets  describe  circles  round  the  sun  with  different 
periodic  times,  such  that  the  squares  of  the  periodic  times  are  proportional  to 
the  cubes  of  the  radii  of  the  circles,  show  that  the  accelerations  of  the  planets 
are  inversely  proportional  to  the  squares  of  their  distances  from  the  sun. 


Vectors 

13.  AVe  have  found  three  kinds  of  quantities,  —  motion,  velocity, 
and  acceleration,  —  all  of  which  can  he  compounded  according  to 
the  parallelogram  law. 

Quantities  which  can  be  compounded  according  to  the  parallelo- 
gram law  are  called  vectors.  A  vector  must  have  magnitude  and 
direction,  and  hence  must  be  capable  of  representation,  on  an 
assigned  scale,  by  a  straight  line.  We  have  seen  that  motion, 
velocity,  and  acceleration  are  all  vectors. 

Composition  and  Resolution  of  Vectors  in  a  Plane 

14.  By  definition  of  a  vector,  two  vectors  can  be  compounded 
into  one,  by  application  of  the  parallelogram  law.  It  also  fol- 
lows from  the  definition  that  any  one  vector  may  be  regarded  as 
equivalent  to  two,  these  two  being  represented  by  the  edges  of  a 
parallelogram  constructed  so  as  to  have  the  original  vector  repre- 
sented by  the  diagonal ;  or,  as  we  shall  say, 
any  vector  can  be  resolved  into  two  others. 

In  particular,  if  we  construct  a  rectangu- 
lar parallelogram  so  as  to  have  a  line  which 
represents  a  vector  R  as  its  diagonal,  we  find 

Fig.  10  ,  ^       ,  ,  f      1    . 

that  the  vector  R  can  be  resolved  uito  two 
vectors  R  cos  e  and  R  sin  e,  at  right  angles  to  one  another,  and  in 

TT 

directions  such  that  R  makes  angles  e,  —  —  e  with  them. 

If  we  take  two  fixed  rectangular  axes  Ox,  Oy  in  a  plane,  we  see 
that  any  vector  R  can  be  resolved  into  two  components  R  cos  e, 


YECTOES  IN  A  PLA^^E 


17 


R  sin  e  parallel  to  these  axes,  where  e  is  the  angle  whicli  It 
makes  with  Ox.  The  components  R  cos  e,  R  sin  e  are  spoken  of 
as  the  comjyonents  of  R  along  Ox  and  Oi/. 

There  are  two  ways  of  compounding  a  number  of  vectors  R^, 
^2>  ■  ■  • '  ^n-  I^  tl^6  first  place,  we  can  construct  a  polygon 
ABODE  -N,  such  that  the  sides  AB,  BC,  CD,  ■■■,  MN  repre- 
sent the  vectors  R^,  R^,  R^,  ■■■,  R^.  Then  AN  will  represent  the 
resultant.  For  R^,  R^  can  first  be  compounded  into  a  vector  R^^ 
represented  by  A  C.  Combining  Rg 
with  this  vector,  we  obtain  a  vec- 
tor represented  by  AD,  and  so  on 
until  finally  AN  is  reached. 

As  a  second  way,  we  can  resolve 
each  vector,  such  as  R^,  into  its 
two  components 

i?^  cos  e^,     R^sine^, 

along   rectangular   axes    Ox,    Oy. 

The  n  vectors  are  now  resolved 

into  2  n  vectors,  of  which  n  are  parallel  to  Ox  and  n  are  parallel 

to  Oy.    The  first  set  of  n  can  be  compounded  into  a  single  vector 

X  =  i?j  cos  Cj  -f  R^  cos  63  4-  •  •  • 
parallel  to   Ox,  while  the  second  set  can  be  compounded  into  a 
single  vector  y  -  i?,  sin  e,  +  i?.  sin  e^  +  •  •  • 

parallel  to  Oy.    We  now  have  two  vectors  A",  Y  parallel  to  Ox,  Oy. 

If  their  resultant  is  a  vector  R  making  an  angle  e  with  Ox,  we 

have  ^  cos  e  =  X  =  R^  cos  e^  +  R^  cos  €„  -f  •  •  •.  (1) 

i?  sin  e  =  Y  =  R^  sin  e^  +  R^shx  €^+  ■■■.  (2) 

To  find  the  numerical  value  of  R,  we  square  and  add  (1)  and 
(2)  and  obtain 

R^  =  x^  +  Y- 

=  {R^  cos  e^  -I-  Ro  cos  €„  +  ■•■y+  (i?i  sin  ej  -|-  R.,  sin  e.,  -I-  •  •  •)' 
=  Rl  +  R;  +  ■■■  +  2 R^R^ (cos  e^  cos  e^  +  sin  e^  sin  e,)  -(-  ■  ■  • 
=  Rl  +  RI  +  ---  +  2  R^R^  cos  (cj  -  e.,)  -f-  •  •  •. 


Fig.  11 


\{oO 


18 


EEST  AND  MOTION 


To  find  the  direction  of  the  resultant,  we  divide  the  correspond- 
ing sides  of  (1)  and  (2)  and  obtain 

Y      R,  sm  e,  +  R^  sin  e,  +  •  ■  • 
tan  e  =  —  =  — ^ 


X      R^  cos e^  +  R^  cos e^  +  ■  ■■ 

If  we  nave  only  two  vectors  R^,  R^,  making  an  angle  6  with 
one  another,  we  may  put  Cj  —  Cg  =  ^>  ^^^  obtain 

R''  =  R\  +  Rl+2  R^R^  cos  6. 

Since  R  is  obviously  tlie  diagonal  of  a  parallelogram  having  two 
edges  of  lengths  R^,  R^,  meeting  at  the  angle  6,  this  result  can 
be  obtamed  du-ectly  from  the  geometry  of  the  triangle  ADC,  in 
which  the  angle  at  C  is  evidently  ir  —  0. 
Thus 

R"  =  R\  +  Rl  -  2  R^R^  cos  (tt  -  6), 

which  is  clearly  identical  with  the  above 
expression. 

We  may  take   two   examples  to   illustrate 
the  method  of  resohnng  vectors  into  rectangular  components  in  a  plane. 

1.  In  Ex.  2,  p.  10,  suppose  that  the  direction  of  the  ship  {AB  in  fig.  5) 
is  taken  for  axis  Ox,  and  that  that  in  which  the  shot  is  to  travel  is  taken 
for  axis  Oy.  Let  the  shot  be  fired  with  velocity  F,  making  an  angle  6  with 
Ox,  the  velocity  of  the  ship  being  v.  The  resultant  velocity  is  to  be  along 
Oy,  so  that  the  velocity  along  Ox,  say  A",  is  to  be  nil.    We  have,  however, 

« 
X  =  V  +  Fcos^, 

V 

so  that  we  must  have  cos  6  = ,  giving 

the  result  already  obtained. 

2.  To  find  'the  acceleration  of  a  jioint 
moving  with  uniform  velocity  V  in  a  circle 
of  radius  a.  Let  A  be  the  position  of  the 
particle  at  time  t  =  0,  and  take  OA  for 
axis  of  X.  After  time  t  the  particle  has 
described  a  length  Vt  of  arc,  so  that  if 
B  is  its  position  after   time  t,   the   angle 

BOA    is     —    in    circular    measure.     The 
a 

direction   of  velocity  at  B,  namely  the  tangent  at  B,  will  accordingly 


VECTORS  IN  SPACE 


19 


Vt 


make  an  angle  -  H with  Ox,  so  that  the  components  of  the  velocity 

along  Ox,  Oy,  say  I'l,  v^,  will  be 

rr        (^       Vt\  ^^   .     Vt 

V,  —  V  cos  I  — I )  —  —  V  sin  —  , 

\2        « /  a 


Vo  =  V  sin 


(i^v)  = 


K  cos  —  • 
a 


di\ 


The  acceleration  along  Ox  is   — ^>  which,  on  differentiating  r^  with 
respect  to  t,  is  found  to  be 

F2        P^< 


•  cos  — , 

a         a 

while  that  along  Oy  is  similarly  found  to  be  — ? ,  or 

(It 
F2   .    Vt 

sm  — 

a  .       a 


72 


Compounding  these,  we  obviously  obtain  an  acceleration  —  along  BO, 
the  residt  already  obtained  on  page  15.  ^ 


z 

E 

/ 

/ 

___ — ■ — ■ 

B 

C 

X 

/■ 

± 

/ 

D 

Composition  and  Resolution  of  Vectors  in  Space 

15.  It  may  be  tliat  the  vectors  to  be  compounded  are  not  all  in 
one  plane.  However,  the  method  of  determining  the  resultant  is 
essentially  the  same.  Thus  we  can  con- 
struct a  polygon  in  space  ABCD  ••■  N 
such  that  the  sides  AB,  BC,  ■■■,  MN rep- 
resent the  vectors  R^,  •■■,  R^.  As  in  the 
preceding  case,  it  is  readily  shown  that 
ylA^is  the  resultant. 

It  is  usually  more  convenient  to  resolve 
each  vector  into  three  components  par- 
allel to  rectangular  axes  in  space.    Given 

a  vector  AB,  we  draw  through  A,  and  likewise  through  B,  three 
planes  parallel  to  the  coordinate  planes.  They  inclose  a  rectangular 
parallelepiped  of  which  AB  is  a  diagonal.  The  edges  AC,  AD, 
AE  represent  three  vectors  by  which  AB  can  be  replaced ;  they 
are  the  components  parallel  to  the  axes  of  the  vector  AB. 

Suppose  there  are  n  vectors,  and  that  the  direction  angles  of 
the  vector  R^  are  denoted  by  a.,  /3^,  7^.    As  above,  each  vector  R^ 


Fig. 14 


20  REST  AND  jSrOTIOX 

can  be  replaced  by  three  components  parallel  to  the  axes;  these 
vectors  are  of  amount 

i?^.  cos  «, ,     i?,  cos  ^, ,     i?,  cos  7, . 
Of  the  Sn  vectors  thus  obtained,  the  n  vectors  parallel  to  the 
a:-axis  can  be  compounded  into  the  smgle  vector 

X  =  H^  cos  a^  +  ^2  cos  a,  +  ••  ■  +  it„  cos  a^.  (3) 

The  whole  system  of  vectors  can  thus  be  replaced  by  tliis  vector 
and  two  others  parallel  to  the  y  and  z  axes  respectively,  namely 

Y=  Pi^  cos  /3i  +  B,  cos  /3,  H h  i?„  cos  /3„.    '         (4) 

Z=  It^  cos  7i  +  7^2  cos  72  ^ +  E^  cos  j„.  (5) 

Evidently  the  resultant  of  these  three  vectors,  and  consequently  of 
the  original  n  vectors,  is  a  diagonal  of  a  rectangular  parallelepiped 
whose  edges  are  of  lengths  X,  Y,  Z.    If  the  length  of  the  resultant 
be  denoted  by  R,  and  the  direction  angles  by  a,  /3,  7,  we  have 
R''  =  X^  +  Y'^  +  Z\ 

X  Y  Z 

and  cos  a  =  —  >     cos  /3  =  —  5     cos  7  =  —  • 

R  R  '       R 

Hence  the  residtant  is  completely  determined  in  magnitude  and 
direction. 

Ccntroids 

16.  Let  a  system  of  vectors  be  represented  in  direction  by  OA^, 

OA^,  •  ■  -,  OA^,  and  let  their  magnitudes  be  mpA^,  ■  •  •,  mjDA^,  where  m^, 

m^,  •  •  ■,  ??!-„  are  any  quantities.    Denote  by  x^,  y^, z^  the  coordinates  of 

A^  with  respect  to  axes  through  0 ;  by  a^,  /3^,  7^  the  direction  angles 

of  OA^  with  respect  to  these  axes ;  and  by  R^  the  magnitude  of  the 

vector  m^OA^.    The  components  of  this  vector  along  these  axes  are 

given  by  j-,  ^  . 

"  -^  R,.  cos  a,.  =  m^OA^  cos  a^  =  m,.x^, 

R^  cos  /3,.  =  m^OA^  cos  /3^  =  m^y^, 

^^  cos  7^  =  m,.OA^  cos  7^  =  m^.z^. 

Hence  equations  (3),  (4),  (5)  can  be  written  thus : 

n  n  n 

X  =  ^m^x^,       Y  =  ^m^y^,      Z  ^^m^z,..     •  (6) 

1  1  1 


m.  Q&GRGE  F.  IVlcEWEN 


THE   CEXTKOID 


21 


For  the  interpretation  of  this  result,  we  make  use  of  the  idea  of 
the  centroid  of  a  system  of  points.  By  definition  the  centroid  of  a 
system  of  points  is  the  point  such  that  its  distance  from  any  one 
of  three  coordinate  planes  is  the  average  of  the  distances  of  all  the 
pomts  of  tlie  system  from  this  plane,  it  bemg  understood  that  each 
distance  is  measured  with  its  proper  algebraic  sign. 

From  tliis  definition,  it  follows  that  the  distance  of  the  centroid 
from  any  plane  whatever  is  equal  to  the  average  of  the  distances 
of  the  n  points  from  this  plane.  For  if  x^,  y^,  z^  are  the  coordinates 
of  the  rth  point,  the  coordinates  of  the  centroid,  say  x,  y,  z,  will  be 


a;  =  —  > 


X, 


y 


Xy,.,    ~^  =  -2- 


(7) 


and  the  perpendicular  distance  from  the  centroid  to  any  plane 

ax  +  hy  -\-  cz  -{-  d  =  0 

1 

(ax  +  hy  +  cz  -\-  d) 


IS 


Va^  +  &==  +  c^ 

_  1  4^  ax^  +  ly^  +  cz^-\-  d 
~  ^  T       Va^  +  1^  +  0^ 

which  proves  the  result. 

Let  us  imagine  that  of  the  n  points  a  number  ??i„  all  coincide 
at  the  point  x^,  y^,  z^,  a  number  m^,  at  the  point  a;^,  y^,  2;^,  and  so  on. 
Then  the  centroid  has  coordinates  (by  equations  (7)), 


n  A 


m  x^  = 


(8) 


22  l^EST  AND  MOTION 

where  the  summation  is  now  taken  over  the  various  points  in  space 
at  which  the  original  points  are  accumulated.  Calling  these  points 
in  space  A,B,C,--,  the  point  x,  y,  z  is  said  to  be  the  centroid  of  the 
points  A,B,C,--,  correspondmg  to  the  multipliers  m„,  m^,  m^,  •  •  •. 
By  means  of  these  results,  equations  (6)  are  reducible  to 

X=S;.^m,,      Y^Tj.^m,.,     Z=z-^m,.  (9) 

1  1  1 

Hence  the  resultant  of  the  above  set  of  vectors  is  directed  along 

n 

the  line  (9C,  and  its  magnitude  is  OC-^m^.    As  defined  by  equa- 

1 
tions  (9),  the  multipliers  m^  can  be  any  numbers  whatever,  positive 

n 

or  negative,  so  that  the  sum  V?^,.  may  be  positive,  zero,  or  nega- 

1 
tive.    In  particular,  when  the  vectors  are  represented  in  magnitude 

as  well  as  direction  by  OA^,  ■•■,  0A„,  the  resultant  is  directed  along 

OCg,  and  its  magnitude  is  w  •  OC^,  where  n  is  the  number  of  vectors 

and  the  point  C'„  is  the  centroid,  as  defined  above.    Thus  we  have 

the  theorem : 

Theoeem.  If  vectors  of  magnitude  m^OA^,  m„OA^,  •••  act  along 
the  lines  OA^,  OA^,  ■■■,  then  their  resultant  is  of  magnitude 
{m^  +  mg  +  •  •  •)0G,  and  acts  along  OG,  where  G  is  the  centroid  of 
the  points  A^,  A^,  •  ■  -for  the  tnultipliers  ?n-p  m^,  ■  •  ■. 

Obviously  the  parallelogram  law  is  a  particular  case  of  this 
theorem. 

EXAMPLES 

1.  Find  the  resultant  of  two  vectors  of  magnitudes  5  P  and  12  P  which  meet 
at  right  angles. 

2.  A  vector  P  is  the  resultant  of  two  vectore  which  make  angles  of  30°  and 
45°  with  it  on  opposite  sides.    How  large  are  the  latter  vectors  ? 

3.  Show  how  to  determine  the  directions  of  two  vectors  of  given  magnitude 
so  that  their  resultant  shall  be  of  given  magnitude  and  direction.  When  is  this 
impossible  ? 

4.  Show  that  if  the  angle  at  which  two  given  vectors  are  inclined  to  each 
other  is  increased,  their  resultant  is  diminished. 

5.  Under  what  conditions  will  the  resultant  of  a  system  of  vectors  of  magni- 
tudes 7,  24,  and  25  be  equal  to  zero  ? 


EXAMPLES  23 

6.  Three  vectors  of  lengths  P,  P,  and  P  V^  meet  in  a  point  and  are  mutu- 
ally at  right  angles.  Determine  the  magnitude  of  the  resultant  and  the  angles 
between  its  direction  and  that  of  each  component. 

7.  Three  vectors  of  lengths  P,  2  P,  3  P  meet  in  a  point-  and  are  directed 
along  the  diagonals  of  the  three  faces  of  a  cube  meeting  at  the  point.  Detennine 
the  magnitude  of  their  resultant. 

8.  Show  that  the  resultant  of  three  vectors  represented  by  the  diagonals  of 
three  faces  of  a  parallelepiped  meeting  in  a  vertex  A  is  represented  by  twice 
the  diagonal  of  the  parallelepiped  drawn  from  A. 

9  i)  is  a  point  in  the  plane  of  the  triangle  ABC,  and  I  is  the  center  of  its 
inscribed  circle.  Show  that  the  resultant  of  the  vectors  a  •  AD,  b  ■  BD,  c  ■  CD  is 
(a  4-  b  +  c)  ID,  where  a,  b,  c  are  the  lengths  of  the  sides  of  the  triangle. 

10.  ABCD,  A'B'C'D'  are  two  parallelograms  in  the  same  plane.  Find  the 
resultant  of  vectors  drawn  from  a  point  proportional  to  and  in  the  same  direc- 
tion as  AA',  B'B,  CC,  D'D. 

11.  If  0  is  the  center  of  the  circumscribed  circle  of  the  triangle  ABC  and 
P  its  orthocenter,  show  that  OP  is  the  resultant  of  the  vectors  OA,  OB,  and 
OC  ;  also  that  2  PO  is  the  resultant  of  PA,  PB,  PC. 

12.  The  chords  AOB  and  COD  of  a  circle  intersect  at  right  angles.  Show 
that  the  resultant  of  the  vectors  OA,  OB,  OC,  OD  is  represented  by  twice  the 
vector  OP,  where  P  is  the  center  of  the  circle. 

GENERAL  EXAMPLES 

(In  these  examples  take  the  acceleration  produced  by  gravity  to  be  32  feet  per  second 

per  second) 

1.  A  point  possesses  simultaneously  velocities  of  2,  3,  8  feet  per  .second, 
in  the  directions  of  a  point  describing  the  three  sides  of  an  equilateral  tri- 
angle in  order.    Find  the  magnitude  of  its  velocity. 

2.  A  point  possesses  simultaneously  velocities,  each  equal  to  r,  in  the 
directions  of  lines  drawn  from  the  center  of  a  regular  hexagon  to  five  of 
its  angular  points.  Find  the  magnitude  and  direction  of  the  resultant 
velocity. 

3.  When  a  steamer  is  in  motion  it  is  found  that  an  awning  8  feet  above 
the  deck  protects  from  rain  the  portion  of  the  deck  more  than  4  feet  behind 
the  vertical  projecti6n  of  the  edge  of  the  awTiing ;  but  when  the  steamer 
comes  to  rest  the  line  of  separation  of  the  wet  and  dry  parts  is  6  feet  in 
front  of  this  projection.  Find  the  velocity  of  the  steamer,  if  that  of  the 
rain  be  20  feet  per  second. 

4.  A  ship  sailing  along  the  equator  from  east  to  west  finds  that  from 
noon  one  day  (local  time)  to  noon  the  next  day  (local  time)  the  distance 
covered  is  420  miles.  What  would  be  the  day's  run,  if  the  ship  were  sail- 
ing at  the  same  rate  from  west  to  east  ? 


24  EEST  AND  MOTION 

5.  A  railroad  nins  due  east  and  west  in  latitude  X.  At  what  rate 
must  a  train  travel  along  the  road  to  keep  the  sun  always  directly 
south  of  it? 

6.  Determine  the  true  course  and  velocity  of  a  steamer  going  due  north 
by  compass  at  10  knots  through  a  4-knot  curi-ent  setting  southeast;  and 
determine  the  alteration  of  direction  by  compass  in  order  that  the  steamer 
should  make  a  true  northerly  course. 

7.  A  bicyclist  rides  faster  than  the  velocity  of  the  wind,  and  makes  the 
ei-ror  of  judging  the  direction  of  the  wind  to  be  the  direction  in  which  it 
appears  to  meet  him  when  he  is  in  motion.  Show  that  the  wind  will 
always  appear  to  be  against  him,  in  whatever  direction  he  rides. 

8.  One  ship  sailing  east  with  a  speed  of  20  knots  passes  a  light- 
ship at  11  A.M. ;  a  second  ship  sailing  south  at  the  same  rate  passes  the 
same  point  at  1  p.m.  At  what  time  are  they  closest  together,  and  what  is 
then  the  distance  between  them  ? 

9.  Two  particles  move  with  velocities  v  and  2  v  respectively  in  oppo- 
site directions,  in  the  circumference  of  a  circle.  In  what  positions  is  their 
relative  velocity  greatest  and  least,  and  what  values  has  it  then? 

10.  Find  the  relative  motion  of  two  particles  moving  with  the  same 
velocity  i',  one  of  which  describes  a  circle  of  radius  a  while  the  other  moves 
along  a  diameter. 

•  11.  Two  particles  move  unifo7-mly  in  straight  lines.  At  a  given  time 
the  distance  between  them  is  a  and  their  relative  velocity  is  V,  the  com- 
ponents of  the  latter  in  the  direction  of  a  and  perpendicular  to  it  being  u 
and  V.  Show  that,  when  they  are  nearest  together,  their  distance  is  av/V, 
and  that  they  arrive  at  this  position  after  the  interval  au/  V'-. 

12.  Three  horses  in  a  field  are  at  a  certain  moment  at  the  vertices  of 
an  equilateral  triangle.  Their  motion  relative  to  a  jierson  driving  along  a 
road  is  in  a  direction  round  the  sides  of  the  triangle  (in  the  same  sense), 
and  in  magnitude  equal  to  the  velocity  of  the  carriage.  Show  that  the 
three  horses  are  moving  along  concurrent  lines. 

13.  Two  points  describe  concentric  circles,  of  radii  a  and  h,  with  speeds 
varying  inversely  as  the  radii.  Show  that  the  relati-ve  velocity  is  paral- 
lel to  the  line  joining  the  points  when  the  angle  between  the  radii  to 

these  points  is 

.    2  aft 


14.  A  stone  dropped  from  a  balloon  moving  horizontally  is  observed  to 
be  4  seconds  in  the  air,  and  to  strike  the  earth  in  a  direction  making  an 
angle  of  15°  with  the  vertical.    Find  the  velocity  of  the  balloon. 


EXAMPLES  25 

15.  A  ball  is  thrown  from  the  top  of  a  building  with  a  velocity  of  64  feet 
per  second  at  an  angle  of  30°  with  the  horizontal  in  an  upward  direction. 
Find  the  directions  of  its  motion  at  the  end  of  the  first  and  second  seconds, 
and  also  the  velocities  at  these  instants. 

16.  A  ball  is  tossed  into  the  air  with  a  velocity  of  20  feet  a  second,  and 
at  the  end  of  a  second  is  seen  to  be  moving  in  a  line  at  right  angles  to  the 
direction  of  projection.    What  is  its  velocity  at  the  instant? 

. — ^17.  If  the  velocity  of  a  bullet  is  supposed  to  be  a  uniform  horizontal 
/  velocity  equal  to  n  times  that  of  sound,  show  that  the  points  at  which  the 
* — J  sounds  of  the  firing  and  of  the  bullet  striking  the  target  are  heard  simul- 
y  taneously  lie  on  a  hyperbola  of  eccentricity  n.  Examine  the  case  in  which 
y    n  is  very  nearly  equal  to  unity. 

—  18.  Assuming  that  the  earth  moves  in  a  circular  orbit  about  the  sun 
with  a  velocity  29.6  kilometers  per  second,  and  that  the  velocity  of  light 
is  300,000  kilometers  per  second,  find  the  apparent  displacement  of  the  sun 
due  to  the  earth's  motion. 

19.  Assuming  that  the  earth  in  a  year  describes  a  circle  uniformly 
about  the  sun  as  center,  that  the  distance  between  the  centers  is  220  radii 
of  the  sun,  and  that  the  radius  of  the  sun  is  108  times  that  of  the  earth, 
find  the  velocity  of  the  vertex  of  the  earth's  shadow,  taking  the  sun's 
radius  as  the  unit  of  space  and  a  year  as  the  unit  of  time. 


CHAPTER   II 
FORCE   AND   THE   LAWS   OF   MOTION 

Newton's  Laws 

17.  The  laws  of  motion,  as  we  have  said,  form  the  material  sup- 
plied by  experimental  mechanics  for  theoretical  mechanics  to  work 
with.    These  laws  have  been  stated  in  compact  form  by  Newton : 

Law  I.  Every  hody  continues  in  its  state  of  rest,  or  of  nniform 
'motion  in  a,  straight  line,  except  in  so  far  as  it  is  compelled  hy 
impressed  force  to  change  that  state. 

Law  II.  The  rate  of  change  of  inomentum  is  proportional  to  the 
impressed  force,  and  taJces  place  in  the  direction  of  the  straight  line 
in  which  the  force  acts. 

Law  III.  To  every  action  there  corresponds  an  equal  and  oppo- 
site reaction. 

18.  These  laws  introduce  several  new  terms,  —  "force,"  "mo- 
mentum," "  action,"  "  reaction,"  —  which  must  be  explained  before 
the  laws  can  be  fully  understood. 

The  first  law  involves  the  idea  of  motion,  which  has  already 
been  discussed,  and  of  force,  which  is  new. 

The  word  "  force "  is  in  common  use.  It  is  associated  in  the 
first  instance  with  muscular  effort ;  for  example,  we  exert  force  in 
pushing  against  an  obstacle.  Scientifically,  however,  the  word  has 
a  wider  use ;  we  say,  for  instance,  that  two  railway  trucks  when 
they  collide  exert  force  on  one  another,  and  that  the  earth  exerts 
force  on  all  bodies,  causing  them  to  fall  towards  it  unless  they  are 
supported  in  such  a  way  that  they  resist  this  force. 

The  first  law  of  motion,  in  fact,  explains  what  is  to  be  understood 
by  force :  it  is  that  which  tends  to  change  the  state  of  rest  of  a 
body,  or  of  uniform  motion  in  a  straight  line. 

20 


THE  FIRST  LAW  27 

Consider,  for  instance,  a  railway  truck  standing  at  rest  on  a  level  line 
of  rails.  If  a  second  truck  runs  into  it  the  first  truck  will  start  into  motion, 
so  that  force  has  been  applied. 

The  first  law,  however,  tells  us  more  than  this.  It  tells  us  that 
if  a  body  is  kept  free  from  the  action  of  forces,  it  will  remain  in 
its  state  of  rest  or  of  uniform  motion  in  a  straight  luie.  Thus  the 
normal  state  for  a  body  to  be  in  is  one  of  rest  or  of  uniform  motion 
in  a  straight  line,  i.e.  motion  with  uniform  velocity ;  it  is  only 
the  presence  of  force  which  can  alter  tliis  normal  state. 

Consider  again  the  case  of  the  railway  truck.  Let  us  suppose  it  has 
been  set  in  motion  by  collision,  and  that  it  starts  off  with  a  velocity  of, 
say,  10  miles  an  hour.  The  first  law  tells  us  that  unless  forces  act  on  the 
truck,  it  will  continue  its  motion  with  an  unaltered  velocity  of  10  miles 
an  hour  in  the  same  straight  line  in  which  it  started.  When  a  truck  is 
actually  started  into  motion  by  collision,  it  may  be  taken  for  granted  that 
it  will  not  continue  in  uniform  motion  in  a  straight  line,  but  will  sooner 
or  later  be  brought  to  rest.  Thus  forces  must  be  at  work.  Let  us  consider 
the  nature  of  these  forces. 

In  the  first  place  there  is  a  force  known  as  the  resistance  of  the  air. 
The  air  in  front  of  the  truck  presses  against  it  in  such  a  way  as  to  retard 
its  motion.  The  air  therefore  exerts  force  on  the  truck  just  as  a  man  might 
exert  force  on  the  truck  by  pressing  against  it  with  his  hand.  This  force 
alone  would  stop  the  truck  in  time. 

Let  us  suppose  that  the  brakes  are  applied,  and  that  the  wheels  are 
gripped  so  firmly  as  to  be  at  rest  relatively  to  the  truck,  so  that  they  slide 
on  the  rails.  There  is  then  a  large  force  applied  to  the  truck  by  the  rails, 
and  this  again  tends  to  stop  the  motion  of  the  truck.  Even  if  the  brakes 
are  not  applied,  so  that  the  w'heels  are  left  free  to  turn,  there  will  still  be  a 
force  applied  by  the  rails,  although  this  force  will  be  smaller  than  before. 

Suppose  that  the  track  is  curved  instead  of  straight.  We  can  imagine 
the  motion  continuing  for  some  time,  but  it  will  be  motion  along  the 
curve,  and  not  motion  in  a  straight  line,  such  as  we  are  told  by  the  first 
law  would  take  place  if  it  were  not  for  the  action  of  force.  Force  has 
therefore  been  ajiplied,  the  force  being  that  of  the  rails  on  the  flanges  of 
the  wheels,  tending  to  turn  the  truck  round  the  curve.  If  the  flanges  were 
not  present,  this  force  could  not  act,  so  that  the  motion  would  continue  in 
a  straight  line  —  the  truck  would  run  off  the  rails. 

As  another  illustration  of  the  meaning  of  the  first  law,  let  us  examine 
the  motion  of  a  bullet  fired  from  a  gun.  Here  the  forces  which  start  the 
motion  are  supplied  by  the  pressure  of  the  powder.  After  the  bullet  has 
left  the  gun,  the  forces  which  act  are  small  compared  with  tliose  which 


28  FORCE  AND  THE  LAWS  OF  MOTION 

have  started  the  motion,  so  that  we  get  an  approximation  to  uniform 
motion  in  a  straight  line.  The  forces  acting  to  alter  this  motion  are 
(a)  the  resistance  of  the  air  and  (b)  the  weight  of  the  bullet.  The  former, 
as  we  have  seen,  tends  to  stop  the  motion  by  pressure  on  the  ends  and  sides 
of  the  bullet ;  the  latter  tends  to  drag  the  bullet  down  to  the  earth,  and 
so  causes  it,  instead  of  describing  a  straight  line,  to  describe  a  path  which 
curves  downward  towards  the  earth. 

19.  The  conception  of  uniform  motion  in  a  straight  line,  or  of 
rest  (the  particular  case  of  uniform  motion  in  wliich  the  velocity 
is  nil),  as  being  the  normal  state  of  a  body  is  due  to  Galileo  (1564- 
1642).  An  interesting  account  of  the  discovery  of  this  law  will  be 
found  in  Chapter  II  of  Mach's  Science  of  Mechanics}  or  in  Chap- 
ter IX  of  Cox's  Mechanics?'  Before  the  time  of  Galileo  it  was 
commonly  supposed,  on  the  authority  of  Aristotle,  that  every  body 
had  a  natural  place,  and  that  its  normal  state  was  one  of  rest  in 
this  natural  place.  For  instance,  a  stone  was  supposed  to  sink  in 
water,  not  because  the  force  of  gravity  was  acting  on  it  and  setting 
it  into  downward  motion,  but  because  its  natural  place  was  at  the 
bottom  of  the  water;  a  cork  was  supposed  to  rise  because  its 
natural  place  was  at  the  top.  Thus  Girard,^  in  1634,  speaks  of 
"  millions  de  matieres,  qui  sont  disposees  chacunes  en  leurs  lieux," 
and  defines  gravity  as  "  la  force  qu'une  matiere  demonstre  a  son 
obstacle,  pour  retourner  en  son  lieu."  Thus  the  effect  of  force, 
before  Galileo,  was  supposed  to  be  to  keep  a  body  out  of  its 
natural  place.  Galileo  perceived  that  bodies  had  no  natural 
-places  at  all,  but  natural  states,  namely  of  rest  or  of  uniform 
motion  in  a  straight  line,  and  the  effect  of  force  is  not  to  move  a 
body  from  its  natural  place  but  to  disturb  it  from  its  natural 
state,  —  i.e.  to  alter  its  speed.  This  discovery  of  Galileo  is  what  is 
expressed  by  Newton's  first  law  of  motion. 

20.  Having  settled  what  is  meant  by  the  natural  state  of  a 
body  and  also  what  is  meant  by  force,  —  namely  that  wliich  tends 

1  Ernest  Mach,  Science  of  Mechanics  (Eng.  trans,  by  McCormack). 

2  J.  Cox,  Mechanics,  Cambridge,  University  Press,  1904. 

3  In  the  Elzevir  edition  of  Stevin,  Leyden,  1034.  See  Cox,  Mechanics,  loc.  cit. 
ante. 


THE  SECOND  LAW  29 

to  alter  the  natural  state,  —  we  next  inquire  as  to  what  is  the  law 
which  governs  the  effect  produced  by  force.  Given  a  force,  by 
how  much  will  this  alter  the  natural  state  of  uniform  motion  in  a 
straight  line  ?    An  answer  to  this  is  provided  by  the  second  law : 

Law  II.  The  rate  of  change  of  momentum  is  proportional  to  the 
impressed  force,  and  takes  ijlace  in  the  direction  of  the  straight  line 
in  ivhich  the  force  acts. 

The  force,  then,  produces  change  in  a  certain  quantity,  —  the 
momentum  of  the  body  on  which  the  force  acts,  —  and  the  force 
is  proportional  to  the  rate  of  change  of  this  momentum. 

By  momentum  is  meant  the  product  of  the  velocity  of  the  body 
by  a  quantity  known  as  the  mass  of  the  body.  The  mass  meas- 
ures simply  the  quantity  of  matter  of  which  a  body  is  composed, 
and  so  does  not  depend  on  the  motion  of  the  body.    Thus 

rate  of  change  of  momentum  =  mass  X  rate  of  change  of  velocity 

=  mass  X  acceleration, 

by  the  definition  of  acceleration.  We  therefore  see  that  the  force 
is  proportional  to  the  product  of  two  quantities,  the  mass  of  the 
body  and  its  acceleration. 

21.  Measurement  of  mass.  If  we  drop  a  body  from  our  hand, 
it  will,  in  general,  be  acted  on  by  two  forces,  the  resistance  of  the 
air  and  its  weight.  If  we  suspend  the  body  in  a  vacuum,  with  an 
arrangement  for  letting  it  drop  at  any  instant  we  please,  we  get 
rid  of  the  resistance  of  the  air,  and  the  only  force  acting  on  the 
body  will  be  its  weight.  Now  if  any  two  bodies  are  suspended 
side  by  side  in  a  vacuum,  and  are  let  fall  at  precisely  the  same 
instant,  it  will  be  found  that  they  remain  side  by  side  during  the 
whole  time  they  are  falling  towards  the  earth.  Thus  at  any 
mstant  their  accelerations  are  the  same. 

It  follows  from  the  second  law  of  motion  that  the  forces  acting 
are  proportional  to  their  masses.  These  forces,  as  we  have  seen,  are 
simply  the  weights  of  the  bodies,  so  that,  as  the  experimental 
result  is  true  whatever  the  two  bodies  may  be,  we  have  the  gen- 
eral law  :   The  masses  of  bodies  arc  proportional  to  their  weights. 


GIFT  OF 

30  ¥^'RT^AXb  THE  LAWS  OF  MOTIOX 

This  gives  us  a  means  of  comparing  the  masses  of  any  two 
bodies.  In  every  country  a  certain  mass  is  taken  as  standard,  and 
the  mass  of  any  other  body  is  then  compared  with  this  standard,  or 
with  a  copy  of  it,  and  in  this  way  we  get  a  knowledge  of  the  actual 
mass  of  a  body.  For  instance,  in  saying  that  the  mass  of  a  body 
is  n  pounds  we  mean  that  its  mass  (or  weight)  is  equal  to  n  times 
the  mass  (or  weight)  of  a  certain  standard  body  kept  at  London. 

22.  Measurement  of  force.  The  weight  of  a  unit  mass  is  a 
force  which  may  conveniently  be  taken  to  represent  a  unit  force, 
and  if  this  is  done  all  other  forces  may  be  compared  with  this 
force.  Thus  a  force  of  m  pounds  weight  will  mean  a  force  m  times 
as  great  as  the  weight  of  the  standard  pound. 

This  unit  of  force,  however,  is  convenient  rather  than  scientific, 
since  it  varies  when  the  mass  is  moved  about  from  place  to  place 
on  the  earth's  surface.  A  unit  pound  mass  will  weigh  more  at 
London  than  at  Washmgton  ;  for  instance,  it  will  be  found  to 
extend  or  compress  the  sprmg  of  a  spring  balance  more  at  London 
than  at  Washington,  so  that  if  the  pound  weight  is  taken  as  unit 
of  force,  we  must  remember  that  the  unit  of  force  is  different  at 
different  parts  of  the  earth's  surface,  and  that  a  force  of  m  pounds 
weight  at  London  will  be  different  from  a  force  of  m  pounds 
weight  at  Washington. 

For  this  reason  a  second  unit   of  force  is  generally  used  for 

scientific  purposes.    This  is  called  the  absolute  unit  of  force,  and 

is  chosen  so  as  to  be  independent  of  position  on  the  earth's  surface. 

The  second  unit  of  force  is  defined  to  be  one  which  produces  unit 

acceleration  in  unit  mass,  whereas  the  former  unit  produced  an 

acceleration  equal  to  the  value  of  gravity  at  the  point.    Thus,  if  g  is 

the  value  of  gravity,  i.e.  the  acceleration  of  a  body  falling  freely  in 

a  vacuum,  the  practical  unit  equals  g  times  the  absolute  unit. 

If  unit  force  produces  unit  acceleration  in  unit  mass,  a  force  P 

p 
will  produce  in  mass  m  an  acceleration  —    Hence,  denoting  the 

m 

acceleration  by  /,  we  have  the  fundamental  equation 

P^mf,  (10) 


THE   THIRD   LAW  31 

which  is  the  mathematical  expression  of  Newton's  second  law. 
Here  the  force  P  must  be  measured  in  aljsolute  units. 

23.  Law  III.  To  every  action  there  corresponds  an  equal  and 
opposite  reaction. 

It  is  a  matter  of  common  observation  that  a  body  A  cannot 
exert  force  on  a  second  body  B  without  B  at  the  same  time  exert- 
ing force  on  A.  Thus  an  athlete  trying  to  throw  the  hammer  has 
to  be  on  his  guard  that  the  hammer  does  not  throw  liim  ;  the  force 
he  exerts  on  the  hammer  is  accompanied  by  the  hammer  exerting 
force  on  him,  and  he  must  steady  himself  against  the  effects  of 
tliis  force.  So  also  when  a  gun  fires  a  shot  by  exertmg  force  on  it, 
the  shot  exerts  force  on  the  gun,  which  is  shown  in  the  recoil  of 
the  gun.  Thus  all  forces  occur  in  pairs,  which  may  conveniently 
be  spoken  of  as  action  and  reaction.  The  third  law  of  motion  tells 
us  that  the  two  forces  which  constitute  such  a  pair  are  equal  in 
magnitude  and  opposite  in  direction. 

The  meaning  of  the  third  law  will  be  seen  on  examining  the  reaction 
corresponding  to  the  forces  wliich  we  have  already  used  for  illustrative 
purposes.  The  first  illustration  employed  was  that  of  a  collision  between 
two  railway  trucks.  Truck  ^-1  runs  into  truck  D,  exerting  force  on  it  and 
setting  it  in  motion.  The  third  law  tells  us  that  at  the  instant  of  collision 
B  must  exert  force  on  A ,  this  force  being  equal  in  amount  to  that  exerted 
by  A  on  B,  and  opposite  in  direction.  The  force  of  reaction  will  result  in 
a  change  of  velocity  of  A,  lasting  during  the  instant  of  collision  only,  and 
this  may  either  merely  check  the  motion  of  A,  so  that  after  the  collision 
A  proceeds  with  diminished  velocity,  or  it  may  reverse  the  motion  of  .1 , 
so  that  truck  A  is  seen  to  rebound  from  B  and  return  in  the  direction  in 
which  it  came. 

After  B  has  been  set  in  motion  we  have  imagined  it  to  be  acted  on  by 
three  forces  : 

(a)  the  resistance  of  the  air; 

(b)  the  friction  of  the  rails ; 

(c)  the  pressure  of  the  rails  on  the  flanges,  turning  the  truck  round  a 
curve. 

The  reaction  corresponding  to  the  first  force  is  a  force  exerted  by  the 
truck  on  the  air  in  front  of  it  and  near  it,  tending  to  set  the  air  in  motion 
in  the  direction  in  which  the  car  is  moving  ;  it  is,  in  fact,  this  force  which 
clears  the  air  away  from  the  space  occupied  at  any  instant  by  the  truck. 


32  FORCE  AXD   THE  LAWS  OF  INIOTION 

The  reaction  corresponding  to  the  second  force  is  a  force  tending  to 
drag  the  rails  along  Avith  the  truck.  The  rails  are,  of  course,  fastened 
down,  so  that  this  force  cannot  actually  jiroduce  motion. 

The  reaction  corresponding  to  the  third  force  is  a  force  exerted  by  the 
flanges  of  the  truck  wheels  on  the  outer  rail  of  the  curve.  The  rails  press  on 
the  flanges  in  a  direction  towards  the  center  of  the  curve,  so  that  the  flanges 
press  the  rails  outwards  and  away  from  the  center  of  the  curve.  If  the  rails 
are  not  securely  fixed,  this  pressure  will  cause  them  to  move  in  the  direction 
just  mentioned  ;  the  rails  will  "  spread"  and  the  truck  will  run  off  the  track. 

In  the  illustration  of  the  bullet  we  again  had  three  forces  operating  on 
the  bullet : 

(a)  the  pressure  of  the  powder  before  the  shot  leaves  the  barrel  ; 

(b)  the  resistance  of  the  air  during  the  flight  of  the  bullet ; 

(c)  the  weight  of  the  bullet,  dragging  it  downwards  to  the  earth. 

The  reaction  corresjjonding  to  the  first  force  is  the  jiressure  of  the  shot 
driving  the  powder  back.  This  in  turn  is  transmitted  to  the  gun,  producing 
the  "  recoil  "  of  the  gun. 

The  reaction  corresjionding  to  the  second  force,  just  as  with  the  truck, 
sets  the  air  in  motion,  carving  out  a  path  for  the  bullet  and  producing 
the  wind  which  accompanies  its  flight. 

The  reaction  corresponding  to  the  third  force,  the  weight  of  the  bullet, 
is  more  interesting,  because  we  can  obtain  no  direct  evidence  as  to  its 
existence.  We  merely  infer  from  the  princijile  of  the  uniformity  of  nature 
that  as,  in  every  case  which  has  ever  been  tested,  an  action  is  accompanied 
by  an  equal  and  opposite  reaction,  therefore  in  this  case,  which  is  similar 
except  in  that  it  cannot  be  tested,  we  may  suppose  the  action  to  be  accom- 
panied by  its  equal  and  opposite  reaction. 

The  force  which  we  can  observe  is  the  weight  of  the  bullet,  dragging 
it  earthwards.  This,  we  believe,  represents  a  force  exerted  by  the  earth 
itself  on  the  bullet,  —  the  force  of  gravitation.  This  force  must  be  accom- 
panied by  its  reaction,  so  that  the  bullet  must  act  on  the  earth  with 
a  force  equal  to  the  weight  of  the  bullet,  this  force  dragging  the  earth 
upwards  to  meet  the  bullet.  The  force  exerted  by  the  bullet  on  the 
earth  is,  by  the  third  law,  just  as  great  as  that  exerted  by  the  earth  on  the 
bullet.  The  upM'ard  acceleration  2:iroduced  in  the  earth  by  the  bullet  is, 
however,  very  much  less  than  the  downward  acceleration  produced  in  the 
bullet  by  the  earth  ;  for  the  force  is  jointly  proportional  to  the  mass  and 
acceleration  of  the  body  acted  on,  and  as  the  mass  of  the  earth  is  very 
great  compared  with  that  of  the  bullet,  its  acceleration  will  be  very  small 
in  comparison  with  that  of  the  bullet. 

Although  for  these  reasons  the  acceleration  jiroduced  in  the  earth  by 
a  bullet  flying  above  it  cannot  be  observed  directly,  yet  in  a  very  similar 
case  the  acceleration  can  be  observed  directly. 


FRAME   OF   PtEFEREXCE  33 

The  moon,  in  describing  a  circle  round  the  earth,  is  believed  to  be  acted 
upon  by  the  earth's  gravitation  in  just  the  same  way  as  the  bullet.  If  no 
force  acted  on  the  moon,  it  would  describe  a  straight  line  ;  as  it  is,  it  is 
continually  dragged  down  towards  the  earth,  as  we  believe,  by  tlie 
same  force  of  gravitation  as  the  bullet.  Just  in  the  same  way,  then,  the 
earth  ought  to  experience  an  acceleration  towards  the  moon.  This  accelera- 
tion is  one  which  admits  of  astronomical  observation. 

24.  In  terms  of  ideas  which  have  now  been  explained,  the  three 
laws  may  be  restated  as  follows : 

I.  The  normal  state  of  a  body  is  one  of  no  acceleration.  De- 
partures from  this  normal  state  are  produced  b}  the  action  of 
force. 

II.  When  a  force  acts  so  as  to  disturb  the  normal  state  of  a 
body,  the  force  is  proportional  to  the  product  of  the  mass  of  the 
body  by  the  acceleration  produced. 

III.  Forces  occur  in  pairs,  every  action  being  accompanied  by  a 
reaction,  and  each  pair  of  forces  being  equal  and  opposite. 

Frame  of  Eeference 

25.  In  stating  the  laws  of  motion  we  have  spoken  of  the  motion 
of  a  body  without  specifying  the  frame  of  reference  relatively  to 
which  this  motion  is  to  be  measured.  In  practice,  motion  is  gen- 
erally measured  relatively  to  the  surface  of  the  earth,  whereas 
Newton  believed  it  to  be  possible  to  imagine  a  frame  of  reference 
actually  fixed  in  space,  and  intended  all  motion  to  be  measured 
relatively  to  tliis  frame.  Thus  Newton's  laws  of  motion  apply  to 
motion  referred  to  axes  fixed  in  space,  whereas  wdiat  we  require 
to  know,  for  all  proljlems  except  those  of  astronomy,  are  the  laws 
of  motion  referred  to  axes  moving  with  the  earth. 

Let  us  first  consider  the  effect  of  referring  motion  to  a  set  of 
axes  moving  with  uniform  velocity  in  a  straight  line  through 
space.  A  body  under  the  action  of  no  forces  will  have  no  accel- 
eration in  space,  and,  therefore,  as  the  axes  themselves  have  no 
acceleration  in  space,  will  have  no  acceleration  relatively  to  the 
movuig  axes.    Again,  an  acceleration  has  the  same  value  whether 


34  FORCE  AND  THE  LAWS  OF  MOTION 

referred  to  axes  fixed  iu  space  or  to  the  moving  axes ;  for  the 
acceleration  referred  to  the  moving  axes  is  obtained  by  compound- 
ing the  acceleration  referred  to  axes  fixed  in  space  with  that  of  the 
moving  axes,  and  this  latter  acceleration  is  nil. 

Thus  it  appears  that  the  laws  of  motion  retain  exactly  the  same 
form  when  the  motion  is  referred  to  axes  which  move  in  space, 
provided  that  these  axes  move  with  no  acceleration. 

This  condition  of  no  acceleration  is  not  satisfied  by  a  set  of 
axes  fixed  in  the  earth's  surface.  A  point  on  the  earth's  surface 
describes,  on  account  of  the  earth's  rotation,  a  circle  about  the 
earth's  axis.  If  a  is  the  radius  of  this  circle,  and  v  the  velocity 
with  which  it  is  described,  the  point  will  have,  by  §  12,  an  accelera- 

tion  —  towards  the  earth's  axis  of  rotation.    Thus  a  set  of  axes 
a 

fixed  in  the  earth's  surface  will  have  an  acceleration  of  this 
amount,  and  this  has  to  be  borne  in  mind  in  applying  the  laws  of 
motion.  At  a  point  on  the  equator  v  =  46,510  centimeters  per 
second  and  a  =  637xl0^  centimeters,  so  that  the  acceleration  is 

1'" 

—  =  3.4  centimeters  per  second  per  second.    A  body  dropped  at  the 

equator  will  appear  to  have  an  acceleration  of  978.1  centimeters  per 
second  per  second,  if  the  motion  is  referred  to  axes  fixed  in  the 
earth ;  but  will  have  a  true  acceleration  of  amount 

978.1  +  3.4  =  981.5, 
if  the  acceleration  is  referred  to  axes  fixed  in  space. 

This  explains  part  of  the  reason  why  the  force  of  gravity  appears 
to  vary  from  point  to  point  at  the  earth's  surface.  The  weight  of 
a  mass  of  one  kilogramme  will  produce  a  certain  extension  of  the 
spring  of  a  spring  balance  at  the  North  Pole.  If  taken  to  the 
equator,  part  of  the  weight  goes  towards  producing  the  acceleration 
of  the  mass  towards  the  earth's  center,  and  it  is  only  the  remain- 
der which  extends  the  spring  of  the  balance.  The  first  part  is  the 
weight  of  about  3^  grammes  ;  the  remainder  is  the  weight  of  aboiit 
996^-  grammes.  Thus  we  may  say  that,  owing  to  the  acceleration 
of  the  earth's  surface  towards  its  center,  a  mass  of  a  kilogramme 


FKAME   OF   FtEFERE:N:CE  35 

at  the  equator  will  appear  to  act  on  a  spring  balance  with  a  force 
equal  only  to  the  earth's  attraction  on  9961-  grammes. 

A  second  set  of  errors  would  be  introduced  by  referring  motion 
to  axes  in  the  earth's  surface,  these  being  caused  by  the  change  in 
the  directions  of  the  axes.  For  instance,  if  we  use  the  laws  of 
motion  as  though  they  were  true  for  motion  referred  to  axes  fixed 
in  the  earth,  and  apply  these  laws  to  the  fall  of  a  stone,  we  shall 
find  that  the  stone  ought  to  strike  the  ground  at  a  point  vertically 
below  that  from  which  it  is  dropped.  If  we  allow  for  the  rotation 
of  the  earth,  we  shall  find  that  the  point  at  which  the  stone  actually 
strikes  must  be  "somewhat  to  the  east  of  the  point  vertically  below 
that  from  which  it  started. 

The  errors  mtroduced  by  treating  motion  on  the  earth  as  though 
it  were  motion  with  reference  to  axes  fixed  in  space  are,  in  gen- 
eral, either  extremely  small  or  very  easily  corrected.  We  shall, 
therefore,  proceed  at  present  by  neglecting  such  errors  altogether, 
and  shall  apply  the  laws  of  motion  to  motion  with  reference  to 
the  earth's  surface. 

Laws  Applicable  only  to  Motion  of  a  Particle 

26.  There  is  a  further  limitation  to  the  completeness  of  New- 
ton's laws  which  ought  to  be  noticed  here.  The  second  law  would 
lead  us  to  suppose  that  from  a  knowledge  of  the  force  acting  on  a 
body,  and  the  mass  of  the  body,  we  could  deduce  a  definite  accel- 
eration of  the  body.  But  if  the  body  is  of  finite  size,  the  accelera- 
tion will  be  different  at  different  points  of  the  body ;  for  example, 
we  have  seen  that,  as  a  consequence  of  the  earth's  rotation,  the 
acceleration  of  a  point  at  the  equator  of  the  earth  is  different  from 
that  of  a  point  at  the  North  Pole.  "Which  acceleration,  then,  is  it 
that  is  determined  by  the  second  law  ? 

The  answer  to  this  difficulty  is  that  the  second  law  must  be 
supposed  to  apply  only  to  particles,  i.e.  to  pieces  of  matter  so 
small  that  they  may  be  regarded  as  points.  A  moving  particle  has 
a  single  definite  acceleration,  just  as  a  moving  point  has.    From  the 


36  FORCE  AND  THE  LAWS  OF  MOTION 

law  as  applied  to  particles  we  shall  be  able  to  deduce  laws  which 
shall  apply  to  bodies  of  finite  size.  This  problem  will  be  treated 
in  a  later  chapter.  Although,  however,  in  strictness,  the  laws 
ought  to  be  applied  only  to  particles,  it  is  obvious  that  there  may 
be  many  problems  in  wliich  we  can  treat  bodies  of  finite  size  as 
particles  without  introducing  any  appreciable  error.  Such  a  case 
occurred  when  we  discussed  the  flight  of  the  bullet  in  §  18 :  the 
size  of  the  bullet  did  not  come  into  the  question,  as  we  could 
imagine  all  the  points  of  the  bullet  to  have  the  same  acceleration. 
Many  cases  will  occur  in  which  a  body  of  finite  size  may  be  treated 
as  though  it  were  a  particle.  In  the  next  chapter  we  shall  con- 
sider the  application  of  forces  to  particles  and  to  bodies  which  we 
find  it  is  permissible  to  treat  as  particles. 


CHAPTER  III 
FORCES  ACTING  ON  A  SINGLE  PARTICLE 

Composition  and  Eesolution  of  Fokces 

27.  The  second  law  of  motion  enables  us  to  find  the  acceleration 
produced  when  a  particle  of  known  mass  is  acted  upon  by  a  known 
force.    In  nature,  however,  forces  do  not  generally  act  singly. 

Consider,  for  example,  the  flight  of  the  rifle  bullet,  discussed  in 
§  18.  While  the  bullet  is  in  the  air  it  is  acted  on  by  its  weight 
and  by  the  resistance  of  the  air  simultaneously.  In  addition  to 
these,  there  may  be  a  cross  wind  blowing  and  acting  on  the  bulkt 
with  a  horizontal  pressure  in  a  direction  perpendicular  to  its 
motion.  The  resistance  of  the  air  retards  the  motion  of  the  bul- 
let, i.e.  produces  an  acceleration  in  a  direction  opposite  to  that  of 
the  bullet's  motion ;  the  weight  of  the  bullet  drags  it  down,  i.e. 
produces  an  acceleration  towards  the  earth ;  while  the  cross  wind 
will  blow  the  bullet  out  of  its  course,  i.e.  vrill  produce  an  accel- 
eration in  the  direction  in  which  the  wind  is  blowing.  Thus  we 
can  regard  the  three  forces  as  each  producing  its  own  acceleration. 
The  three  accelerations  can  each  be  calculated  from  the  second 
law  of  motion,  and  on  compounding  these  three  accelerations  we 
shall  have  the  resultant  acceleration  of  the  bullet.  This  resultant 
acceleration  could  have  been  produced  by  the  action  of  a  certani 
single  force,  so  that  we  may  say  that  this  single  force  is  equiva- 
lent, as  regards  the  acceleration  produced,  to  the  combination  of 
.the  three  separate  forces,  or  that  the  single  force  is  the  resultant 
of  the  three  separate  forces. 

We  must  now  put  these  ideas  into  exact  mathematical  form. 
As  a  preliminary,  let  us  notice  that  a  force  has  magnitude  and 
direction,  so  that  it  can  be  represented  by  a  straight  line.  We  shall 
show  that  forces  may  be  compounded  accordmg  to  the  parallelogram 

37 


38         FORCES  ACTING  ON  A  SINGLE  P ARTICLE 

law.  Having  proved  tliis,  it  will  follow  that  forces  are  vectors, 
and  may  be  resolved  and  compounded  according  to  the  general 
rules  already  given. 

28.  Parallelogram  of  forces.  Theorem.  If  two  forces  are 
represented  in  magnitude  and  direction  hy  the  two  sides  of  a  par- 
allelogram, their  resultant  will  he  represented  hy  the  diagonal  of 
the  p)(^')^(^ll<^logra')n. 

Let  AB,  AC  represent  the  two  forces,  and  let  Ah,  Ac  represent 
the  accelerations  they  would  produce  if  they  acted  on  any  particle 

separately.  Since,  by  the  second  law 
of  motion,  the  acceleration  is  propor- 
tional to  the  force,  we  must  have 

Ah  :  Ac  =  AB  :  AC. 

Construct  the  parallelograms  Ahdc, 
ABDC.    On  account  of  the  proportion 
just  obtained,  the  two  parallelograms 
will  be  similar,  so  that  AdD  will  be  a  straight  line,  and  we  shall  have 

AD:Ad  =  AB  :  Ah. 

But  Ad,  the  diagonal  of  the  parallelogram  of  edges  Ah,  Ac,  repre- 
sents the  resultant  acceleration.  Since  AB  represents  the  force 
necessary  to  produce  acceleration  Ah,  it  follows  from  the  proportion 
just  obtained  that  AD  will  represent  the  force  necessary  to  produce 
acceleration  Ad.  In  other  words,  the  acceleration  of  the  particle 
is  the  same  as  if  it  were  acted  on  by  a  single  force  represented  by 
AD.    Thus  AD  represents  the  resultant  of  the  forces  AB,  AC. 

It  now  follows  that  force  is  a  vector,  so  that  forces  can  be 
compounded  according  to  the  laws  explained  in  §§  14-16. 

Particle  in  Equilibrium 

29.  In  statics  we  are  concerned  only  with  particles,  or  systems 
of  particles,  at  rest.  The  resultant  force  on  each  particle  must 
accordingly  be  nil.  It  is  therefore  important  to  consider  cases  in 
which  the  resultant  of  a  system  of  forces  is  nil. 


COXDITIONS  FOR  EQUILIBRIUM  39 

30.  Polygon  of  forces.  Theorem.  If  forces  acting  on  a  particle 
are  represented  hij  stravjlU  lines,  the  particle  will  be  in  equilihrium 
if  the  polygon  formed  hy  taking  all  these  straight  lines  as  edges  is 
a  closed  polygon,  i.e.  if  after  p)utting  all  the  lines  end  to  end  ive 
come  back  to  the  starting  point. 

Let  AB,  BC,  CD,  ■  ■  ■,  MN  represent  in  magnitude  and  direction 
any  number  of  forces  which  act  simultaneously  on  a  particle. 
Since  force  is  a  vector  the  forces  represented  by  AB  and  BC  are 
equivalent  to  a  single  force  repre- 
sented by  AC,  and  may  therefore 
be  replaced  by  this  force. 

Thus  the  system  of  forces  may 
now  be  supposed  to  be  forces  repre- 
sented by  the  lines  AC,  CD,  ■  ■  ^MN. 
The  first  two  of  these  may  again 
be  replaced  by  a  single  force  repre- 
sented by  AD,  so  that  the  system 
is  reduced  to  forces  represented  by 
AD,  DE,  •  •  • ,  MN.    We  can  proceed  ^'''-  ^^ 

in  this  way  until  we  are  left  with  only  a  single  force  represented 
by  AN.    This  therefore  represents  the  resultant  of  all  the  forces. 

If  the  polygon  is  a  closed  polygon,  the  points  A  and  ^coincide, 
so  that  the  resultant  force  represented  by  AN  vanishes  and  the 
particle  is  in  equilibrium.  Conversely,  if  the  particle  is  in  equilib- 
rium, AN  vanishes,  so  that  the  polygon  is  a  closed  polygon.  Thus 
the  condition  for  equilibrium  expressed  by  the  theorem  just  proved 
is  necessary  and  suficient,  —  necessary  because  the  condition  must 
be  satisfied  if  the  particle  is  to  be  in  equilibrium,  and  suihcient 
because  equilibrium  is  insured  as  soon  as  the  condition  is  satisfied. 

31.  Triangle  of  forces.  If  there  are  only  three  forces,  the  theo- 
rem reduces  to  a  simpler  theorem  known  as  the  triangle  of  forces. 
This  is  as  follows : 

Theorem.  If  a  particle  is  acted  on  by  three  forces  represented 
by  straight  lines,  the  particle  will  be  in  equilibrium  if  these  three 
straight  lines  placed  end  to  end  form  the  sides  of  a  triangle. 


40         FORCES  ACTING  ON  A  SINGLE  PARTICLE 

As  this  is  a  particular  case  of  the  polygon  of  forces  no  separate 
proof  is  needed.  As  before,  the  converse  is  also  true,  so  that  the 
condition  is  a  necessary  and  sufficient  condition  for  equilibrium. 

When  there  are  only  three  forces  acting,  the  condition  for  equi- 
librium can  be  expressed  in  a  still  simpler  form : 

^  V   32.  Lami's   Theoeem.     When   a  particle  is  acted  on  hy  three 
'       forces,  the  necessary  and  sujficient  condition  for  equilibrium  is  that 
the  three  forces  shall  he  in  one  plane  and  that  each  force  shall  he 
pro2Jortioual  to  the  sine  of  the  a^igle  hetiveen  the  other  two. 

Suppose  that  a  particle  is  acted  on  by  three  forces  F,  Q,  E.   The 
necessary  and  sufficient  condition  for  equilibrium  is  that  we  can 
form  a  triangle  by  placing  end  to  end  three 
lines  which  represent  the  forces  P,  Q,  II  in 
magnitude  and  direction. 

Let  us  begin  by  taking  AB  to  represent 
P,  and  placing  against  it  at  7?  a  line  BC  to 
represent  Q.  Then  CA  must  represent  E,  if 
the  conditions  for  equilibrium  are  to  be  satis- 
tied.  Thus  the  three  forces  must  be  in  one 
plane,  namely,  the  plane  parallel  to  ABC 
through  the  point  of  action  of  the  forces. 

Assume  that  there  is  equilibrium,  so  that 

the  three  forces  are  represented  by  the  sides  of 

the  triangle  ABC.    Let  us  denote  the  angles  of  the  triangle  as  usual 

by  A,  B,  C,  and  its  sides  by  a,  h,  c.    Then,  from  a  known  property 

of  the  triangle,  7 

°    '  a      _      0  c 

sin  A      sin  B      sin  C 

By  our  construction,  however,  a,  h,  c  are  proportional  to  the 
magnitudes  of  the  forces :  we  have 

c        h  _  a 

f^e^q' 

Thus  ^  ^  ^ 


sin  C      sin  A      sin  B 


CONDITIONS   FOR  EQUILIBKIUM  41 

If  pq  denote  the  angle  between  the  lines  of  action  of  the  forces 
P  and  Q,  we  have  jjq  =  ir  —  B,  so  that  sin  B  =  sin  2'>q,  and  hence 

F  Q  H 

~. =  -^^!^—  =  -. (11) 

sm  qr      sin  rp      sin  ptq 

The  converse  is  true  because,  if  the  relation  (11)  is  satisfied  and 
if  the  lines  of  action  of  the  forces  are  in  one  plane,  we  can  con- 
struct a  triangle  of  which  the  sides  will  represent  the  forces  P,  Q,  B, 
so  that  there  is  equilibrium. 

33.  Analytical  conditions  for  equilibrium.  Expressed  in  an  ana- 
lytical form,  the  condition  for  equilibrium  is  that  the  resultant  of  all 
the  forces  acting  shall  be  zero.  If  the  individual  forces  are  known, 
the  resultant  force  can  be  obtained  at  once  from  the  rules  for  the 
composition  of  vectors,  which  have  already  been  given  in  §§  14—16. 

If  the  forces  all  act  in  a  plane,  let  their  magnitudes  be  B^,  B^,  ■•  -, 
-K„,  and  let  their  lines  of  action  make  angles  e^  e^,  ■■  ■,  e„  with  the 
axis  of  X.    Then  the  resultant  has  components  X,  Y,  where  (cf.  §  14) 

X  =  B^  cos  e^  +  i?2  cos  e^  +  ■  ■  ■, 
Y  =  B^  sin  e^  +  B^  sin  e.,  +  •  ■  • . 


The  magnitude  of  the  resultant  is  Va"^+  Y'^,  and  this  vanishes 
only  if  X  and  Y  vanish  separately.  Thus  the  condition  for  equi- 
librium is  that  the  components  along  the  two  axes  shall  vanish 
separately,  i.e.  that  the  sum  of  the  components  of  the  separate 
forces  acting  shall  vanish  when  resolved  along  each  axis. 

Similarly,  if  the  forces  do  not  all  act  in  one  plane,  the  condition 
for  equilibrium  is  that  the  sums  of  the  components  along  three 
axes  in  space  shall  vanish  separately. 

EXAMPLES 

1.  Forces  of  12  and  8  pounds  weight  act  in  two  directions  which  are  at  right 
angles.    Find  the  magnitude  of  their  resultant. 

2.  Three  forces  each  equal  to  F  act  along  three  rectangular  axes.  Find  their 
resultant. 

3.  The  resultant  of  two  forces  Pj  and  Po  acting  at  right  angles  is  R  :  if  Pi 
and  P2  be  each  increased  by  3  pounds,  R  is  increased  by  4  pounds  and  is  now 
equal  to  the  sum  of  the  original  values  of  Pi  and  Po.    Find  Pi  and  Po. 


42  FORCES  ACTING  ON  A  SINGLE  PARTICLE 

4.  Forces  acting  at  a  point  0  are  represented  by  OA,  OB,  OC,  •  •  ■ ,  ON.  Show 
that  if  they  are  in  equilibrium,  O  is  the  centroid  of  the  points  A,  B,  C,  ■  •  ■ ,  N. 

5.  ABCDEF  is  a  regular  hexagon.  Find  the  resultant  of  the  forces  repre- 
sented by  AB,  AC,  AD,  AE,  AF. 

6.  ABCDEF  is  a  regular  hexagon.  Show  that  the  resultant  of  forces  repre- 
sented by  AB,  2 AC,  ZAD,  4:AE,  5AF  is  represented  by  V351  •  AB,  and  find 
its  direction. 

7.  ABC  is  a  triangle,  and  P  is  any  point  in  BC.  If  PQ  represent  the  resultant 
of  the  forces  represented  by  AP,  PB,  BC,  show  that  the  locus  of  Q  is  a  straight 
line  parallel  to  BC. 

Types  of  Forces 
Weight  of  a  Particle 

34.  The  weight  of  a  particle  acts  always  vertically  down- 
ward ;  for  at  a  given  place  on  the  earth  it  is  found  that  the 
weights  of  all  particles  act  in  parallel  directions,  and  this  direc- 
tion is  called  the  vertical  at  the  place  in  question.  The  weight  is 
the  gravitational  force  with  which  the  particle  is  attracted  by  the 
earth,  except  for  a  small  correction  which  has  to  be  introduced 
on  account  of  the  fact  that  axes  fixed  in  the  earth  do  not  move 
without  acceleration.  This  correction  we  shall  not  discuss  here. 
When  the  weight  of  a  body  is  said  to  be  W,  it  is  meant  that  to 
keep  the  body  at  rest  relatively  to  the  earth's  surface  a  force  W  is 
required  to  act  vertically  upward. 

Tension  of  a  String 

35.  A  string  or  rope  supplies  a  convenient  means  of  applying 
force  to  a  body,  and  this  force  is  spoken  of  as  the  tension  of  the 

J.      g  string.    Let  ABCD  •  •  •  be  the   string, 


■P» — j_  '  B  '  c   '  D  '  E        ^^'^  ^^^  P  be  a  particle  tied  to  the 

string  at  its  end.    Let  the  divisions  AB, 

Fig.  18  ^  ' 

BC,  ■■■  of  the  string  be  so  small  that 
each  may  be  regarded  as  a  particle. 

There  will  be  three  forces  acting  on  any  particle  such  as  BC: 
first,  its  weight;  second,  a  force  exerted  on.  BC  by  the  particle  CD 
of  the  string ;  and  third,  a  force  exerted  on  BC  by  the  particle  AB. 


STRINGS  43 

Generally  the  weight  of  a  string  is  very  slight  compared  with 
the  other  weights  in  the  problem.  It  is  therefore  convenient  to 
regard  a  string  as  having  no  weight  at  all.  In  this  case  there  are 
only  two  forces  acting  on  the  particle  AB,  so  that  for  equilibrium 
these  must  be  equal  and  opposite. 

36.  Flexibility.  A  string  is  said  to  be  perfectly  flexible  when 
the  force  exerted  by  one  particle  on  the  next  is  in  the  direction 
joining  the  two  particles.  Thus,  if  the  string  now  under  discussion 
is  perfectly  flexible  and  weightless,  the  forces  acting  on  the  parti- 
cle BC  are  along  the  directions  ijq,  qr.  To  hold  BC  in  equilibrium 
these  must  be  equal  in  magnitude.  Let  T  be  taken  as  the  magni- 
tude of  each.  Also  the  two  forces  must  be  in  opposite  directions, 
so  that  pqr  must  be  a  straight  line. 

Since  action  and  reaction,  by  the  third  law,  are  equal  and  oppo- 
site, the  force  exerted  by  ^  C  on  CD  must  also  be  T  in  the  direc- 
tion qr.  This  must,  for  equilibrium,  be  equal  and  opposite  to  the 
force  exerted  by  BE  on  CD.  This  force  must  accordingly  be  of 
amount  T,  and  qrs  must  be  a  straight  line. 

We  can  continue  in  this  way,  and  find  that  all  the  particles 
must  lie  in  a  straight  line  pqrs  ■  ■  ■,  and  that  each  acts  on  the  next 
with  the  same  force  T.  Also  the  particle  A  at  the  end  of  the  string 
acts  on  P  with  this  same  force  T  in  the  direction  of  the  string. 
The  force  T  is  called  the  tension.    Thus  we  have  the  following : 

The  tension  of  a  string  at  any  point  P  is  defined  as  the  force 
with  which  the  particle  of  the  string  on  the  one  side  of  P  acts  on 
the  particle  on  the  other  side  of  P. 

The  tension  is  the  same  in  magnitude  and  direction  at  every  point 
of  a  perfectly  flexible,  weightless  string  acted  on  by  no  external  forces. 

Hence  it  follows  that 

A  perfectly  flexible,  iveightless  string  acted  on  by  no  external 
forces  must  be  in  a  straight  line  when  in  equilibrium. 

If  the  tension  vanishes,  there  is  equilibrium  whatever  the  direc- 
tion of  the  elements  of  length  pq,  qr,  ■■■ .  When  the  tension  van- 
ishes the  string  is  said  to  be  unstretched.  Clearly  an  unstretched 
string  can  rest  in  equilibrium  in  any  shape. 


44  FORCES  ACTING  ON  A  SINGLE  PARTICLE 

It  will  be  proved  later  that  when  a  perfectly  flexible,  weightless 
string  passes  over  a  smooth  peg  or  pulley,  the  tension  has  the  same 
magnitude  at  all  points  of  the  strmg,  and  at  points  of  contact  with  the 
peg  or  pulley  its  direction  is  along  the  tangent  to  the  peg  or  pulley, 

37.  If  the  string  is  not  absolutely  weightless,  but  is  very  light,  any 
])article  such  as  q  will  be  acted  on  by  three  forces,  —  its  weight  vertically 
down,  and  the  two  forces  from  the  adjacent  particles  acting  along  jx], 

rq.  By  Lami's  theorem,  each  force 
must  be  proportional  to  the  sine 
of  the  angle  between  the  remain- 
ing two  forces.  Since  the  weight 
Pj(j  j9  is  small,  sin  pqr  must  be  small ;  i.e. 

jyqr  must  be  very  nearly  a  straight 
line.  The  line  cannot  be  perfectly  straight,  however,  unless  the  string  is 
absolutely  weightless  ;  thus  in  a  real  string  there  must  always  be  a  certain 
"  sag,"  due  to  the  weight  of  the  string,  although  this  sag  may  be  so  slight 
as  to  be  imperceptible. 

38.  Extensible  and  inextensible  strings.  The  tension,  as  will 
have  been  seen,  is  a  force  acting  at  every  pomt  of  the  string,  and 
tending  to  stretch  the  string  in  the  direction  of  its  length.  The 
string  either  may  or  may  not  yield  to  this  tendency  to  stretch.  A 
string  which  stretches  under  tension  is  called  extensible ;  a  string 
which  does  not  stretch  at  all,  or  which  stretches  so  little  that  the 
amount  of  stretching  is  inappreciable,  is  called  inextensible. 

Thus  an  inextensible  string  remains  of  the  same  length  what- 
ever tension  is  applied  to  it,  wlide  the  length  of  an  extensible  string 
depends  on  its  tension. 

In  1660  Hooke  discovered  a  law  which  expressed  a  relation 
between  the  tension  and  the  amount  of  stretching  in  a  string :  the 
one  is  proportional  to  the  other. 

Definition.  The  length  of  a  string  lolien  the  tension  is  zero  is 
called  the  "  nattiral  length  "  of  the  string. 

Definition.  The  amount  hy  which  the  length  of  a  stretched 
string  exceeds  the  statural  length  of  the  same  string  is  called  the 
"  extension  "  of  the  string. 

Hooke's  Law.  The  tension  of  a  string  is  projwrtional  to  the 
extension. 


STRINGS  45 

Although  Hooke  discovered  this  Law  in  1000,  he  did  not  publish  it  until 
1070,  and  then  only  in  the  form  of  the  anagram  ceiinosssttuv. 

In  1078  he  explained  that  the  letters  of  the  anagram  were  those  of  the 
Latin  words  ut  tensio  sic  vis,  —  "the  power  of  any  spring  is  in  the  same 
proportion  with  the  tension  thereof."  By  tension  (tensio)  Ilooke  meant 
the  quantity  which  we  have  called  the  "  extension  ";  by  the  power  (m) 
he  meant  the  force  tending  to  stretch  the  sj^ring,  i.e.  the  tension. 

39.  Ilooke's  law  only  enables  us  to  compare  tlie  extensions 
produced  by  different  tensions.  To  find  the  actual  extension  pro- 
duced by  a  given  tension  we  must  know  that  })roduced  by  some 
other  tension  for  comparison. 

Definition.  The  force  required  to  stretch  a  string  to  douhle  its 
natural  length  is  called  the  modulus  of  elasticity  of  the  string. 

Thus  if  a  is  the  natural  length  of  a  string,  and  X  the  modulus 
of  elasticity,  we  know  that  a  tension  X  produces  an  extension  a, 
so  that  a  tension  T  produces  an  extension  Ta/X 

When  we  say  that  a  string  is  inextensible,  we  mean  that  X  is  so 
large  that  the  extension  Ta/X  may  be  neglected. 

Hooke's  law  only  holds  within  certain  limits.  If  we  go  on 
increasing  the  tension  in  a  string  indefinitely,  we  find  that,  after  a 
certain  limit  is  passed,  Hooke's  law  ceases  to  be  true,  and  when  a 
certain  still  greater  tension  is  reached  the  string  -breaks  in  two 
parts. 

EXAMPLES 

1.  A  weight  IF  hangs  by  a  string  and  is  pushed  aside  by  a  horizontal  force 
until  the  string  makes  an  angle  of  4-5°  with  the  vertical.  Find  the  horizontal 
force  and  the  tension  of  the  string. 

2.  A  weight  suspended  by  a  string  is  pushed  sideways  by  a  horizontal  force. 
Show  that  as  it  is  pushed  farther  from  its  position  of  rest,  in  which  the  string 
is  vertical,  the  tension  continually  increases. 

3.  A  weight  of  100  pounds  is  suspended  by  two  strings  which  make  angles 
of  G0°  with  the  vertical.    Find  their  tensions. 

4.  A  weight  of  30  pounds  is  tied  to  two  extensll)le  strings  of  natural  length 
2  feet,  modulus  of  elasticity  100  pounds,  and  the  other  ends  of  the  strings  are 
tied  to  two  points  at  a  horizontal  distance  4  feet  apart.  Find  the  position  in 
which  the  weight  can  rest  in  equilibrium. 

5.  A  weight  W  is  suspended  by  three  equal  strings  of  length  I  from  hooks 
which  are  the  vertices  of  a  horizontal  equilateral  triangle  of  side  a.  Find  the 
tensions  of  the  strings. 


46  FORCES  ACTING  ON  A  SINGLE  PARTICLE 

Reaction  hchvcen  Tivo  Bodies 

40.  A  second  way  in  which  force  can  be  applied  to  a  particle  is 
by  the  pressure  between  the  particle  and  the  surface  of  a  solid 
body.    Such  a  force  is  commonly  spoken  of  as  a  reaction. 

A  body  standing  on  the  floor  of  a  room  is  acted  on  by  its 
weight  acting  downwards,  but  is  kept  at  rest  by  the  action  of  a 
second  force  acting  upwards  from  the  floor;  this  is  the  reaction 
between  the  body  and  the  floor.  Clearly,  in  order  that  the  body 
may  rest  in  equilibrium,  the  reaction  in  this  case  must  be  equal  to 
the  weight  of  the  body  and  must  act  vertically. 

Friction 

41.  Imagine  a  small  body  standing  on  a  plane  of  which  the 
slope  can  be  varied,  such  as  the  lid  of  a  desk.  If  the  plane  is  held 
horizontally,  the  body  can  stand  at  rest  as  already  described.  Let 
the  plane  be  gradually  tilted,  and  it  will  be  found  that  as  soon 
as  the  tdting  reaches  a  certain  angle  the  body  wdl  begin  to  slide 
down  the  plane.  The  angle  at  which  sliding  first  occurs  is  found 
to  be  different  for  different  pairs  of  substances ;  thus  for  wood 
sliding  on  wood  it  may  vary  from  10°  to  25°,  for  iron  on  wood  it 
varies  from  10°  to  30°,  wliile  for  iron  sliding  on  iron  it  is  only 
about  10°  or  15°. 

When  two  substances  are  such  that  this  angle  is  zero,  —  i.e.  such 
that  one  can  only  rest  on  the  other  when  the  surface  of  contact  is 
perfectly  horizontal,  —  then  the  contact  between  them  is  said  to  be 
perfectly  smooth.  The  nearest  approximation  to  a  perfectly  smooth 
contact  which  we  experience  in  actual  life  is  probably  that  of  steel 
on  ice,  as  in  skating. 

It  is  found  that  the  angle  to  which  a  plane  made  of  one  sub- 
stance has  to  be  tilted  before  a  second  substance  begins  to  slide 
on  it  is  uidependent  both  of  the  amount  of  the  second  substance 
and  of  the  area  in  contact.  Thus  the  angle  depends  only  on  the 
nature  of  the  two  substances  in  contact. 


FRICTIOX 


47 


Further,  when  the  two  bodies  are  pressed-  together  in  any 
way  it  is  found  that  the  direction  of  the  reaction  can  make  any 
angle  up  to  a  certain  limiting  angle  with  the  normal  to  the 
plane  of  separation  without  sliding  taking  place,  but  that  as  soon 
as  this  angle  is  reached  sliding  takes  place.  This  angle  is 
known  as  the  angle  of  friction.  It  is  clearly  the  same  as  the 
angle  through  which  the  plane  before  considered  can  be  tilted, 
for  the  angle  between  the  normal  to  the  plane  and  the  direction 
of  the  reaction  (i.e.  the  vertical)  is  simply  the  slope  of  the  plane. 

42.  In  any  case  in  which  frictional  forces  act,  let  R  denote  the 
normal  component  of  the  reaction,  and  let  F  denote  the  compo- 
nent in  the  plane  of  the  contact  which 
is  caused  by  friction.  When  slipping  is 
just  about  to  occur,  the  resultant  must 
make  an  angle  e  with  the  normal,  where 
€  is  the  angle  of  friction.  Thus,  if  S  de- 
notes the  whole  reaction,  we  must  have 

li  =  S  cos  e,     F  =  S  sin  e, 
and  hence        F  —  Fv  tan  e.  Fig.  20 

The  quantity  tan  e  is  called  the  coeflcient  of  friction  and  is  denoted 
by  the  single  letter  fM.  Then,  when  slipping  is  just  about  to  take 
place,  w^e  have  ,-,         ^, 

^         '  I  =  /Xll. 

It  must  be  clearly  understood  that  this  equation  gives  the  true 
value  of  the  frictional  force  onl//  when  slipping  is  just  alont  to 
take  place.  It  sets  an  upper  limit  to  the  value  of  the  frictional 
force,  but  does  not  give  the  actual  value  of  this  force  unless  we 
know  that  the  system  is  on  the  verge  of  sliding. 

43.  Consider,  for  instance,  the  experiment  already  discussed,  in 
which  a  particle  is  placed  on  a  horizontal  plane  wliich  is  gradually 
tilted  up.  When  the  plane  is  horizontal  the  particle  is  at  rest, 
acted  on  only  by  gravity  and  the  reaction  with  the  plane.  Thus 
the  reaction  is  vertical,  so  that  here  F  =  0.  Consider  next  the 
state  of  things  when  the  plane  makes  an  angle  a  with  the  horizon. 


48 


FORCES  ACTING  ON  A  SINGLE  PARTICLE 


Wcos  a 


If  slipping  does  not  take  place,  the  particle  is  in  equilibrium 
under  its  weight  W  and  the  reaction  between  it  and  the  plane. 
Thus  the  reaction  must  consist  of  a  vertical  force  W.    We  can 

resolve  this  into  components 
W  cos  a  and  JF  sin  a  perpen- 
dicular to  and  up  the  plane. 
The  former  is  the  normal  com- 
ponent of  the  reaction,  the 
latter  is  the  frictional  com- 
ponent. Thus  in  the  notation 
already  used  we  have 
B  =  W  cos  a, 
F  =  W  sin  a, 

so  that  in  this  case  we  have 
Fig.  21  F  =  R  tan  a. 

As  a  increases,  F  and  F/B  both  increase  until,  when  a  reaches 
the  value  e,  F/R  reaches  its  limiting  value  fx,  or  tan  e,  and  after 
this  slipping  takes  place. 


EXAMPLES 

1.  A  mass  of  100  pounds  placed  on  a  rough  horizontal  plane  is  on  the  point 
of  starting  into  motion  when  acted  on  horizontally  by  a  force  equal  to  the 
weight  of  100  pounds.    Find  the  angle  of  friction. 

2.  A  body  placed  on  an  inclined  plane  which  makes  an  angle  of  30°  with  the 
horizontal  is  just  on  the  point  of  moving  down  the  plane  when  acted  on  by  a 
horizontal  force  equal  to  the  weight  of  the  body.    Find  the  coefficient  of  friction. 

3.  A  man  capable  of  exerting  a  pull  of  200  pounds  tries  to  drag  a  mass  of 
700  pounds  over  a  horizontal  road  (coefficient  of  friction  \).  To  help  him,  the 
chain  from  a  crane  is  attached  to  the  mass,  the  chain  hanging  vertically.  How 
much  tension  must  there  be  in  the  chain  before  the  man  can  move  the  block  ? 

4.  An  insect  tries  to  crawl  up  the  inside  of  a  hemispherical  bowl  of  I'adius  a. 
How  high  can  it  get,  if  the  coefficient  of  friction  between  its  feet  and  the  bowl  is  ^  ? 

5.  A  man  trying  to  push  a  block  of  stone  over  ice  pushes  horizontally  and 
finds  that  just  as  soon  as  the  stone  begins  to  move  his  feet  begin  to  slip.  Show 
that  if  he  pushes  upwards  on  the  stone  he  will  get  it  along  without  difficulty,  but 
that  if  he  pushes  downwards  he  cannot  possibly  move  it. 

6.  A  smooth  pulley  is  placed  at  the  edge  of  a  horizontal  plane.  A  string 
passes  over  it,  having  at  one  end  a  weight  w  hanging  freely,  and  at  the  other 


ILLUSTRATIVE  EXAMPLES 


49 


end  a  weight  W  resting  on  tlie  plane.  If  tlie  coefficient  of  friction  fj.  is  so  large 
that  motion  does  not  take  place,  find  through  what  angle  the  plane  mast  be 
tilted  before  motion  begins. 

7.  A  tourist  of  mass  M  is  roped  to  a  guide  of  mass  m  on  the  side  of  a  moun- 
tain, the  side  of  which  may  be  talcen  to  be  an  inverted  hemisphere.  The  length 
of  the  rope  subtends  an  angle  a  at  the  center  of  the  mountain,  and  the  rope  is 
not  supposed  to  touch  the  mountain  at  any  point.  If  the  coefficient  of  friction 
between  either  man  and  the  mountain  is  /x,  how  far  can  the  tourist  venture 
down  the  side  of  the  mountain  before  both  he  and  the  guide  fall  to  the  bottom  ? 


ILLUSTRATIVE  EXAMPLES 

1.  A  heavy  particle  C  rests  on  a  smooth  inclined  plane,  being  supported  by 
two  strings  of  lengths  li,  I2,  which  are  attached  to  two  points  A,  B  in  the  plane, 
these  points  being  in  the  same  horizontal  line  and  at  a  distance  h  apart.  Find  the 
tensions  of  the  strings  and  the  reaction  with  the  plane. 

Let  W  be  the  weight  of  the  particle  and  let  ex  be  the  inclination  of  the  plane  to 
the  horizon.    The  particle  is  in  equilibrium,  being  acted  on  by  the  following  forces : 

(a)  Its  weight  W,  which  acts  vertically  downwards. 

(6)  The  reaction  between  the  particle  and  the  plane.  Since  the  plane  is 
smooth,  this  reaction  acts  at  right  angles  to  the  plane.  Let  the  amount  of  the 
reaction  be  R. 

(c)  The  two  tensions  of  which 
the  amount  is  required.  Let 
the  amounts  of  these  be  denoted 
by  Ti,  To. 

Since  these  four  forces  pro- 
duce equilibrium,  the  sum  of 
their  resolved  parts  in  any  di- 
rection must  vanish.  The  two 
tensions  have  no  resolved  parts 
at  right  angles  to  the  plane  ; 
hence,  by  resolving  at  right  an- 
gles to  the  plane,  we  shall  get 
an  equation  in  which  only  two  of  the  forces  are  involved. 

The  resolved  part  of  the  weight  at  right  angles  to  the  plane  is  IF  cos  a.    Tlu' 

reaction  is  wholly  at  right  angles  to  the  plane  ;  hence  the  equation  for  whit'h 

we  are  in  search  is  ^       „^  „ 

R  —  W  cos  a  —  0. 

This  gives  us  the  amount  of  the  reaction  at  once. 

Let  us  now  consider  the  resolved  parts  of  the  forces  in  the  inclined  plane. 
The  only  forces  which  have  components  in  this  plane  are  the  following  : 

(a)  The  weight,  of  which  the  component  is  Tl'sin  a.  down  the  line  of  greatest  slope. 

(b)  The  tensions  of  the  strings,  which  are  entirely  in  the  plane  and  which 
act  along  the  strings  CA,  CB. 


Fig.  22 


50 


FORCES  ACTING   OX  A  SINGLE  PARTICLE 


The  three  forces  IF  sin  a,  Ti,  and  T^  must  be  in  equilibrium;  lienre,  by 
Lami's  theorem,  each  must  be  proportional  to  the  sine  of  the  angle  between 
the  other  two. 

In  fig.  23,  CD,  CA,  CB  are  the  lines  of  action  of  these  three  forces.  The 
line  CD,  being  the  line  of  greatest  slope  through  C,  is  at  right  angles  to  the 
line  AB,  which  we  are  told  is  horizontal.    Thus  if  DC  is  produced  to  meet  AB 

in  P,  the  angle  APC  is  a  right  angle.    Hence 

sin  ACD  =  sin ^CP  =  cos  CAP, 
and  similarly 

sin  BCD  =  sin  BCP  =  cos  CBP. 
By  Lami's  theorem  we  have 

TFsin  a  _        Ti       _        Tg 
sin  AC  B  ~  sin  BCD  ~  sin  ACD 

From  the  relations  just  obtained  we  can 
obtain  these  ratios  in  terms  of  the  angles  of 
the  triangle  ABC.    We  have 


Fig.  23 


IF  sin  a 
sin  C 


Ti 


cos  B      cos  A 


We  can  now,  if  required,  express  cos  J.,  cos  B,  and  sin  C  in  terms  of  the  sides 
li,  I2,  and  h  of  the  triangle,  by  means  of  the  ordinary  formulae  of  trigonometry. 

The  student  is  advised  to  exarhine  for  himself  the  form  assumed  by  the  result 
in  the  two  special  cases 

(a)  A  =  B  =  0,  in  which  ACB  are  in  a  straight  line  ; 

It 
(6)  C  =  0,  ^  =  2?  =  — ,  in  which  the  strings  are  parallel. 

2.  Tmd.  the  direction  and  magnitude  of  the  smallest  force  which  will  start  a 
body  resting  on  an  inclined  plane  into  motion  down  the  plane. 

Let  a  be  the  angle  of  the  plane,  and  n  the  coefficient  of  friction  between  it 
and  the  body  to  be  moved.  Let  W  be  the  weight  of  the  body,  and  let  a  force  F 
be  applied  in  a  direction  making 
an  angle  0  with  tlie  line  of  greatest 
slope  down  the  plane,  this  force 
being  supposed  to  be  just  sufficient 
to  move  the  body. 

The  forces  acting  on  the  body 
consist  of 

(a)  its  weight  W ; 

(6)  the  applied  force  F ; 

(c)  the  reaction  with  the  plane. 

Let  the  last  force  be  resolved  into  two  components  along  and  perpendicular 
to  the  plane.  Taking  the  latter  to  be  R,  the  former  will  be  /xR  acting  up  the 
plane,  for,  by  hypothesis,  the  body  is  on  the  point  of  motion  down  the  plane. 


Fig.  24 


^ 


ILLUSTRATIVE  EXAMPLES 


51 


The  resultant  of  all  these  forces  vanishes,  so  that  the  sum  of  their  com- 
ponents in  any  direction  vanishes.    Resolving  normal  to  the  plane,  we  obtain 

n  +  Fsm e  -  >rcos  a  =  o, 

and  resolving  along  the  plane, 

F  cos  e  +  W  sin  a  —  nR  =  0. 

Eliminating  the  unknown  reaction  E,  we  obtain 

F{fi  sin  d  +  cos  6)  —  W{fi  cos  a  —  sin  a)  =  0, 
_  W{fx  cos  a  —  sin  a) 


so  that 


/n  sin  5  +  cos  6 


Replacing  /j.  by  tan  e,  we  obtain 

_  TF(cos  a  tan  e  —  sin  a)  _  l'rsin(e  —  a) 
cos  6  +  tan  e  sin  6  cos  {d  —  e) 

The  value  of  jF  is  a  minimum  when  cos  {6  —  e)  is  a  maximum,  and  this  occure 
when  cos  (^  —  e)  =  1 ;  i.e.  when  ^  =  e.    In  this  case  the  value  of  F  is 

F=  TFsin(e-  a). 

Thus  this  is  the  smallest  force  by  which  motion  can  be  produced,  and  it  must 
act  so  as  to  make  an  angle  6  with  the  plane  equal  to  the  angle  of  friction  e. 
Since,  by  hypothesis,  the  weight  rests  without  slipping  when  no  force  is  applied, 
the  angle  e  must  be  greater  than  a.  Thus  the  direction  of  the  force  F  must 
always  be  inclined  in  an  upward  direction.  The  function  performed  by  the 
force  F  is  twofold :  it  supports  part  of  the  weight  of  the  body  (through  its 
component  normal  to  the  plane),  and  so  lessens  the  amount  of  friction  to  be 
overcome ;  and  it  also  supplies  (through  its  component  in  the  inclined  plane) 
the  motive  power  for  overcoming  the  frictional  resistance.  When 
these  two  parts  of  the  force  are  balanced  in  the  most  advantageous 
way  the  value  of  F  is  a  minimum,  and  this,  as  we  have  proved, 
occurs  when  d  =  e. 

An  interesting  and  instructive  solution  of  this  problem  can 
also  be  obtained  geometrically.  For  equilibrium,  the  three  forces 
already  enumerated  must  satisfy  the  condition  of  forming  a  tri- 
angle of  forces. 

Let  AB  represent  the  weight  and  BC  the  reaction  between  the 
mass  and  the  plane,  then  CA  must  represent  the  applied  force  F. 
If  the  body  is  on  the  point  of  motion,  the  reaction  must  make  an 
angle  e  with  the  normal  to  the  plane,  so  that  the  angle  ABC  must 
be  e  —  (X.  Thus  the  line  BC  is  fixed  in  direction,  and  the  problem 
is  that  of  finding  the  direction  and  magnitude  of  AC,  when  the  length  -4C  is  a 
minimum.  Obviously  the  minimum  occurs  when  -4  C  is  perpendicular  to  BC,  so 
that  AC  must  be  in  a  direction  making  an  angle  e  —  a  with  the  horizontal,  as 
already  found,  and  since  AC  =  AB  sin  ABC  =  AB  sm{€  —a),  the  magnitude  of 
the  force  required  will  be  TFsin(e  —  a). 


Fig. 


52 


FORCES  ACTING   ON  A  SINGLE  PARTICLE 


3.  A  particle  is  tied  to  an  elastic  string,  the  other  end  of  lohich  is  fixed  at  a 
■point  in  a  rough  inclined  plane.  Find  the  region  of  the  pdane  loithin  which  the 
particle  can  rest. 

The  forces  acting  on  the  particle  are 

(a)  its  weiglat,  say  TT,  vertically  down  ; 

(b)  the  tension  of  the  string ; 

(c)  the  reaction  with  the  rough  plane. 

Let  the  natural  length  of  the  string  be  /,  the  modulus  of  elasticity  \ ;  then 

when  the  actual  length  of  the 

string  is  r,  where  r  is  greater 

(r-Z)X 


than  I,  tlie  tension  is 


I 


Fig.  26 


Let  a  be  the  inclination  of 
the  plane,  and  let  /j.  be  the 
coefficient  of  friction  between 
the  plane  and  the  particle.  Let 
the  reaction  with  the  plane  be 
resolved  into  a  normal  component 
R,  and  a  component  F  along  the 
plane.  The  condition  that  the 
particle  can  remain  at  rest  is 
that  F  shall  be  less  than  fjiR. 
Resolving  at  right  angles  to  the  plane,  the  only  forces  which  have  compo- 
nents in  this  direction  are  found  to  be  the  weight  of  the  particle  and  its  reaction 

with  the  plane.    Thus 

R  --  W  cos  a  =  0. 

Consider  the  equilibrium  of  the  particle  when  at  some  point  P,  distant  r(>l) 
from  O.    The  components  in  the  inclined  plane,  of  the  forces  acting  on  it,  are 
(a)  W  sin  a  down  the  line  of  greatest  slope  through  P ; 

{b)  the  tension —  along  PO  ; 

(c)  the  frictional  component  of  the  reaction,  which  we  have  called  F. 

Let  OP  make  an  angle  d  with  the  line  of  greatest  slope  in  the  plane.  •  Then, 
since  the  resultant  of  the  first  two  forces  must  be  of  magnitude  F,  we  must  have 

(r  —  A2X2       2(r  —  l\\ 
F2  =  wz  sii,2  a  +  x_Jj! —  +     ^     .     '     W  sin  a  cos  6, 


Z2. 


I 


giving  the  magnitude  of  the  frictional  force  required  to  maintain  equilibrium. 
If  the  particle  is  on  the  point  of  motion,  F  =  fiR  =  fiW cos  a,  so  that 

(r  —  7V-2X2       '^(r  —  IW 
W^  (sin2  a-  ii"^  cos2  a)  + —  +  -^ '—  W  sin  a  cos  61  =  0.  (a) 


Since  r,  6  are  polar  coordinates  of  the  point  P,  equation  (a)  is  the  polar  equa^ 
tion  of  the  boundary  of  the  region  within  which  the  particle  can  remain  at  rest. 


ILLUSTRATIVE  EXAMPLES 


h6 


The  equation  is  most  easily  interpreted  by  noticing  tliat  li  r  —  I  is  replaced 
by  r,  the  equation  becomes 

X"-^  X 

W^  (sin2  a  —  ix^  cos'-  cr)  -| r^  +  2  ,^  W  sin  a-  •  r  cos  ^  =  0,  {h) 

which  is  the  polar  equation  of  a  circle.  Thus  the  original  locus  represented  by 
equation  (a)  can  be  drawn  by  first  drawing  the  circle  represented  by  equation 
(6),  and  then  producing  each  radius  vector  through  the  origin  to  a  distance  I 
beyond  the  circumference  of  this  circle. 


Fig.  27 


Fig.  28 


The  same  result  can  be  obtained  by  a  geometrical  treatment  of  the  problem. 
The  particle  is  acted  on  by  only  three  forces  in  the  i^lane  on  which  it  rests,  so 
that  lines  parallel  and  proportional  to  tliese  forces  must  form  a  triangle  of  forces. 

In  fig.  29  let  OP  be  the  string,  and  let  AP  be  a 
length  I  mea.sured  off  from  P,  .so  that  ^0  is  the 
extension  r  —  loi  the  string.  The  tension  is  always 
proportional  to  ^0  and  acts  along  AO.  Let  us 
then  agree  that  in  the  triangle  of  forces  the  tension 
shall  be  represented  by  the  actual  line  AO.  On 
the  same  scale  let  the  component  of  the  weight, 
W  sin  flr,  be  represented  by  the  line  OG,  the  direc- 
tion of  this  being,  of  course,  down  the  line  of  greatest 
slope  through  O.  Then  AOG  must  be  the  triangle 
of  forces,  so  that  GA  must  represent  the  frictional 
reaction  between  the  particle  and  the  plane.  The 
maximum  value  possible  for  this  is  /ilF  cos  a,  .so  that 
if  slipping  is  just  about  to  occur,  GA  will  represent 
a  force  /u  W  cos  a.  Thus  corresponding  to  a  position 
of  P  in  which  slipping  is  just  about  to  occur,  the  positions  of  A  are  such  that  GA 
represents  the  constant  force  iiW  cos  a, — in  other  words,  the  locus  of  ^4  is  a  circle 
of  center  G.    This  leads  at  once  to  the  construction  previously  obtained. 


Fig.  2<) 


54 


FORCES  ACTING   ON  A  SINGLE  PARTICLE 


The  region  in  wliicli  equilibrium  is  possible  assumes  two  different  forms 
according  as  the  angle  of  the  inclined  plane  a  is  less  or  greater  than  the  angle 
of  friction  e.  In  the  former  case  the  region  of  equilibrium  is  of  the  kind  repre- 
sented in  fig.  27.  On  passing  the  value  a  =  e  the  circle  used  in  the  construction 
passes  through  the  point  O,  and  for  values  of  a  greater  than  e  the  region  of 
equilibrium  becomes  an  area  of  the  kind  drawn  in  fig.  28.  On  passing  through 
the  value  a  =  e  a  sudden  change  takes  place  in  the  shape  of  the  region  of 
stability.  For  values  of  a  which  are  greater,  by  however  little,  than  e,  a  circle 
of  radius  I,  center  O,  is  entirely  outside  the  region  of  stability ;  while  for  values 
of  a  which  are  smaller,  by  however  little,  than  e,  this  circle  is  inclosed  within 
the  region  of  equilibrium.  Clearly  this  circle  maps  out  the  region  within  which 
the  weight  can  rest  with  the  string  unstretched,  and  this  will  be  one  of  equi- 
librium or  not  according  as  a  <  or  >  e. 

Thus  this  circle  falls  inside  or  outside  the  region  of  equilibrium  in  the  way 
predicted  by  analysis.  At  the  same  time  we  could  not  have  been  sure,  without 
a  separate  investigation,  that  the  result  given  by  analysis  would  be  accurate  as 
regards  the  region  within  a  distance  I  of  0.  For  the  analysis  began  by  assuming 
the  string  to  be  stretched,  and  so  had  no  application  except  to  the  region  at  a 
distance  greater  than  I  from  0. 


4.  Two  weights  to,  w'  rest  on  a  smooth  sphere,  being  supported  by  a  string 
which  passes  through  a  smooth  ring  at  0,  a  point  vertically  above  the  center  of  the 
sphere.    Find  the  configuration  of  equilibrium. 

Let  P,  Q,  in  fig.  30,  be  the  positions  of  the  two  weights 
in  a  configuration  of  eqviilibrium.  The  weight  lo  at  P  is 
acted  on  by  the  following  forces : 

(a)  its  weight  w  vertically  downwards ; 

(5)  the  tension  of  the  string  along  PO; 

(c)  the  reaction  between  the  sphere  and  the  weight. 
Since  the  sphere  is  supi^osed  smooth,  the  direction  of 
this  reaction  is  at  right  angles  to  the  plane  of  contact 
between  the  particle  and  the  sphere;  i.e.  along  CP. 

The  three  forces  acting  on  the  particle  P  are  accord- 
ingly parallel  to  the  three  sides  of  the  triangle  OPC. 
Thus  the  triangle  OPC  may  be  regarded  as  a  triangle  of 
forces  for  these  forces,  so  that  the  magnitudes  of  the 
forces  must  be  proportional  to  the  sides  of  this  triangle. 
Denoting  the  tension  and  reaction  by  T  and  E,  we  obtain 


Fig.  30 


w   _    T   _   R 

'dc~op~'cp' 


(a) 


In  the  same  way  the  triangle  OCQ  may  be  regarded  as  a  triangle  of  forces  for 
the  particle  Q.  (This  triangle  does  not  represent  force  on  the  same  scale  as  the 
former  triangle  OCP ;  for  in  the  former  case  OC  represented  a  weight  w,  whereas 
it  now  represents  a  weight  w'.) 


ILLUSTRATIVE  EXAMPLES  55 

From  this  second  triangle  of  forces  we  obtain 

w'  _   T'  _  R^  ,j^, 

oc~'oq~'cq' 

where  T',  R'  represent  the  tension  and  reaction  acting  on  Q. 

Since  the  ring  at  O  is  supposed  to  be  smooth,  tlie  tension  in  tlie  string  POQ 

is  the  same  at  all  points.    Thus  T  =  T'.    We  now  obtain,  from  equations  (a) 

and  (6), 

w-  OP  =  w'  ■  OQ,  (c) 

since  each  product  is  equal  to  T  ■  OC.    If  I  is  the  whole  length  of  the  string, 
we  have 

~dq~'op~      I     ' 

showing  that  the  string  arranges  itself  so  that  it  is  divided  by  the  ring  at  O 

in  the  inverse  ratio  of  the  two  weights.    We  notice  also  from  equations  (a) 

and  (h)  that 

R      R' 


for  each  ratio  is  that  of  the  radius  of  the  sphere  to  OC.    Thus  the  reactions  are 
in  the  direct  ratio  of  the  weights. 

If  the  string  is  inextensible,  the  length  I  is  known,  so  that  equations  (d) 
determine  the  lengths  OP,  OQ  completely.  Suppose,  however,  that  the  string  is 
an  extensible  string,  say  of  natural  length  a  and  modulus  X.  Then,  instead  of  I 
being  a  known  quantity,  we  have  one  additional  equation  between  unknown 
quantities,  namely 

r  =  ^^x. 

a 

Remembering   that  equation  (c)   gives  two  values   for   the   quantity   T-  OG, 
we  have 

T-OC  =  ^-^ \.  OC  =  w-OP  =  w'-  OQ 


so  that \  •  00  = 

a 


This  equation  determines  the  value  of  I,  and  having  found  this  we  proceed 
as  before. 

5.  A  weight  W  is  supported  by  strings  of  ickich  the  tensions  are  Ti,  T2,  ■  ■  •  T„. 
The  strings  do  not  hang  vertically,  but  the  angles  between  the  different  pairs  of 
strings  are  known,  being  eio,  eis,  etc.  Find  the  weight  W  in  terms  of  the  tensions 
and  of  these  angles. 


OP  _0Q_ 

I 

1     ~    1    ~ 

w         10' 
I 

1+i 

IV        w 

1        1 

W         10 

56  FORCES  ACTING  ON  A  SINGLE  PARTICLE 

Clearly  the  weight  is  equal  to  the  resultant  of  the  tensions  Ti,  T^,  •  •  •  Tn. 

Let  us  take  any  three  rectangular  axes  in  space.  Let  the  direction  cosines 
of  the  first  string  be  li,  nii,  ni ;  let  those  of  the  second  string  be  l.>,  »i2,  no  ;  and  so 
ou.    Then,  resolved  along  the  axes,  the  components  of  the  first  tension  will  be 

liTi,    niiTu    niTi. 

The  components  of  the  other  tensions  are  similar  expressions,  so  that  if 
X,  Y,  Z  denote  the  three  components  of  the  resultant,  we  have 

X  =  hTi    +I2T.2    +----\-lnT„, 
Y=miTi  +  m.To  +  ■■■  +  ?n„r„, 
Z  =  niTi  +  nsTa  +  •  •  •  +  ?i„r„. 

Since  the  magnitude  of  the  resultant  is  equal  to  11',  we  have 
W^  =  .Y2  +  F2  +  Z2 

=  (hTi  +  hT.2  +  ---  +  InTJ'  +  {miTi  +  mnT2  +  ■■■  +  m„T„y^ 

+  (mTi  +  noTo  +  ■  ■  ■  +  ii„r„)2 
=  Tf  (if  +  m{  +  n{)  +  •  •  ■  +  2  Ti To  (1^1.  +  mini.  +  mno)  +  •  •  • 
=  T{  +  T.f  +  ---  +  2TiT2  cos  €12  +  •  •  •, 

which  gives  the  result  required. 


GENERAL   EXAMPLES 

1.  ABC  is  a  triangle,  with  a  right  angle  at  ^1;  AD  is  the  perpendicular 

on  BC.    Prove  that  the  resultant  of  forces,  acting  along  AB  and  — :^ 

1  A  B  A  O 

acting  along  A  C,  is  — r  acting  along  ^1 D. 
AB 

2.  At  a  point  0  there  acts  a  force  P,  whose  line  of  action  is  in  the  plane 
determined  by  two  lines  OA ,  OB,  meeting  at  O.  The  resolved  part  of  P  in 
the  direction  OA  is  represented  in  magnitude  and  direction  by  OA',  that 
in  the  direction  OB  by  OY.  Show  that  the  force  P  is  represented  in  mag- 
nitude by  the  diameter  of  the  circle  OXY,  and  find  its  direction. 

3.  Forces  P^,  P^,  •  •  -,  P„  acting  in  one  plane  at  a  point  0  are  in  equi- 
librium. Any  transversal  cuts  their  lines  of  action  in  points  L^,  Lc,,  ■  ■  ■,  L„; 
and  a  length  OL,-  is  considered  positive  when  the  direction  from  O  to  Li  is 
the  same  as  to  P^-.    Prove  that  ^P^/OL,  =  0. 

4.  A  body  is  sustained  on  a  smooth  inclined  plane  by  two  forces,  each 
equal  to  half  the  weight,  the  one  acting  horizontally,  and  the  other  along 
the  plane.    Find  the  inclination  of  the  plane. 

5.  The  angle  of  a  smooth  inclined  plane  is  30°,  and  a  force  P  acting 
horizontally  sustains  a  body.  In  what  other  direction  can  P  act  and  sup- 
port the  body  ?    Compare  the  pressure  upon  the  plane  in  the  two  cases. 


EXAMPLES  57 

6.  Two  smooth  planes,  whose  inclinations  are  a  and  /3,  meet  in  a  hori- 
zontal line  AB.  At  a  point  in  AB  is  a  small  smooth  ring  through  which 
passes  a  string  with  a  weight  at  either  en;l,  resting  one  on  each  of  the 
given  planes,  and  in  the  same  vertical  plane  with  the  ring.  If  the  weights 
are  in  the  equilibrium,  find  the  tension  of  the  string  and  the  ratio  of  the 
weights. 

7.  Two  smooth  rings  of  weights  W^  and  Wo  are  connected  by  a  string 
and  rest  in  equilibrium  on  the  convex  side  of  a  circular  wire  in  the  vertical 
plane.  Show  that,  if  the  string  subtends  the  angle  a  at  the  center  of  the 
circle,  the  angle  of  inclination  6  of  the  string  to  the  vertical  is  given  by 

n\  +  W.,        a 

tan  0  = cot  —  • 

Tl\  -  W^         '2 

8.  Two  weights  rest  on  a  rough  inclined  plane  and  are  connected  by  a 
string  which  passes  over  a  smooth  peg  in  the  plane  ;  if  the  angle  of  inclina- 
tion a  is  greater  than  the  angle  of  friction  e,  show  that  the  least  ratio  of 
the  less  to  the  greater  is  sin  (a  —  e)/sin(a  +  e). 

I  .-.''9-  Two  weights  supjjort  one  another  on  a  rough  double  inclined  plane, 
'VMjy  means  of  a  fine  string  passing  over  the  vertex,  and  both  weights  are 
^    about  to  move.    Show  that  if  the  plane  be  tilted  until  both  weights  are 

again  on  the  point  of  motion,  the  angle  through  which  the  plane  will  be 

turned  is  twice  the  angle  of  friction. 

■*' 
1    ^/^lO.  Two  weights  P,  Q  of  similar  material,  resting  on  a  double  inclined 

^^  plane,  are  connected  by  a  fine  string  passing  over  the  common  vertex,  and 

Q  is  on  the  point  of  motion  down  the  plane.    Prove  that  the  greatest  weight 

which  can  be  added  to  P  without  disturbing  the  e(piilibrium  is 

P  sin  2  e  sin  (^r  +  /3) 

; ' 

sin(a  —  6)sin(/3  —  e) 

a,  p  being  the  angles  of  inclination  of  the  planes,  and  e  the  angle  of 
frictibn. 
r\^\.Xil-  A  body  is  supported  on  a  rough  inclined  plane  by  a  force  acting 
■  along  it.  If  the  least  magnitude  of  the  force,  w-hen  the  plane  is  inclined 
at  an  angle  a  to  the  horizon,  be  equal  to  the  greatest  magnitude,  when  the 
plane  is  inclined  at  an  angle  /3,  show  that  the  angle  of  friction  is  §(a  — /3). 

^  12.  Two  equal  rings  of  weight  W  are  movable  along  a  curtain  pole,  the 
coefficient  of  friction  being  yu.  Tlie  rings  are  connected  by  a  loose  string 
of  length  /,  which  supports  by  means  of  a  smooth  ring  a  weight  Wi-  How 
far  apart  must  the  rings  be  so  that  they  will  not  come  together? 

13.  Two  weights  P,  Q  of  different  material  are  laid  on  a  rough  plane, 
whose  inclination  is  6,  and  connected  by  a  taut  string  inclined  at  45°  to  the 
intersection  of  the  plane  Avith  the  horizon.    Both  weights  are  on  the  point 


58  rOECES  ACTIXG   OX  A   SIXGLE  PARTICLE 

of  motion.     Determine  the  coefficients  of  friction  of  P  and  Q,  it  being 
known  tliat  that  of  the  upper  weight  is  twice  that  of  the  lower. 

14.  A  heavy  ring  is  free  to  slide  on  a  smooth  elliptic  W'ire  of  eccen- 
tricity e  in  a  vertical  plane,  the  major  axis  of  the  ellipse  making  an  angle  a 
with  the  horizontal,  and  a  string  fastened  to  the  ring  2>asses  over  a  smooth 
peg  at  the  center  of  the  ellipse  and  supports  a  body  of  equal  weight.  Show 
that  the  angle  </>  which  the  tangent  to  the  wire  at  the  ring  makes  with  the 
major  axis  is  given  by  the  equation 

tan  (0  +  a)  (sec^  (p  —  e'^)  =  e^  tan  </>. 

15.  Two  small  smooth  rings  of  weights  W,  W  are  connected  by  a  string, 
and  slide  on  two  fixed  wires,  the  former  of  which  is  vertical  and  the  latter 
inclined  at  an  angle  a  to  the  horizontal.  A  weight  P  is  tied  to  the  string, 
and  the  two  portions  of  it  make  angles  6,  <t>  with  the  vertical.    Prove  that 

cot  e  :  cot  0  :  cot  a  =  W  ■.P  +  W:P  +  W  +  W. 

16.  Two  particles  of  unequal  mass  are  tied  by  fine  inextensible  strings 
to  a  third  particle.  They  lie  on  a  rough  horizontal  plane,  with  the  strings 
stretched  and  making  angles  a,  /3  with  the  horizontal  line  in  the  plane. 
Find  the  magnitude  and  direction  of  the  least  horizontal  force  which,  on 
being  applied  to  the  third  particle,  will  move  all  three. 

17.  A  heavy  particle  is  placed  on  a  rough  inclined  plane  of  which  the 
inclination  a  is  equal  to  the  angle  of  friction.  A  thread  is  attached  to 
the  particle,  and  passed  through  a  hole  in  the  plane  which  is  lower  than 
the  particle,  but  is  not  in  the  line  of  greatest  slope  through  it.  Show  that 
if  the  thread  be  gradually  drawn  through  the  hole,  the  particle  will  describe 
a  straight  line  and  a  semicircle  in  succession. 


jili^^iJl^'' 


CHAPTEE  IV 


STATICS   OF   SYSTEMS  OF    PARTICLES 


44.  So  far  we  have  been  considering  the  action  of  forces  on  a 
single  particle.  A  different  class  of  problems  arises  in  considering 
the  action  of  forces  on  a  body  composed  of  a  great  number  of  par- 
ticles, to  which  forces  are  applied  in  such  a  way  as  to  act  on  the 
different  particles  of  the  body. 

Consider  what  happens  when  a  force  F  is  applied  to  one  parti- 
cle ^  of  a  body  which  is  composed  of  a  great  number  of  particles 
A,  B,  C,  D,  ■■.  If  the  particle  A 
were  in  no  way  influenced  by  the 
other  particles  B,  C,  D,  ■■  the 
particle  A  would  start  into  motion 
under  the  action  of  the  applied 
force,  and  would  soon  become  sepa- 
rated from  the  other  particles  B, 
C,  D,---.    If,  lio  wever,  the  particles 

A,  B,  C,  £>,■■•  constitute  a  single 
continuous  body,  this  does  not 
happen.  "Wliat  happens  is  that  as 
soon  as  the  particle  A  begins  to  move  relatively  to  the  other  parti- 
cles, systems  of  actions  and  reactions  come  into  play  between  the 
particle  A  and  the  adjacent  particles  B,  C,  J),--.  Speaking  loosely, 
we  may  say  that  the  forces  acting  on  A  tend  to  check  the  motion 
of  A,  while  the  corresponding  reactions  tend  to  impart  motion  to 

B,  C,D,--.  When  B,  C,D,--  start  into  motion,  further  systems  of 
forces  begin  to  operate  on  the  particles  next  beyond  B,  C,  D,  ■  ■, 
and  so  on.  Thus  all  the  particles  are  set  mto  motion,  and  instead 
of  the  particle  A  moving  singly  the  complete  body  moves  as  a  whole. 
We  have  now  to  discuss  whether  such  a  body,  or  system  of  bodies, 

59 


Fig.  31 


60  STATICS  or  syste:\[s  of  particles 

will  move  or  will  remain  at  rest,  when  systems  of  forces  act  from 
outside  on  its  different  particles.  We  shall  have  to  remember 
throughout  that  the  forces  applied  from  outside  are  not  the  only 
forces  acting,  but  that  these  are  accompanied  by  actions  and  reac- 
tions between  the  different  particles. 

45.  One  consequence  of  this  last  fact  appears  at  once.  Applying 
a  force  to  one  particle  ^  of  a  body  is  not  the  same  thing  as  apply- 
ing an  exactly  similar  force  to  another  particle  B.  For  the  systems 
of  internal  actions  and  reactions  will  be  different  in  the  two  cases. 
Any  simple  example  will  show  that  the  resulting  motion  will,  in 
general,  also  be  different ;  e.g.  a  horizontal  force  applied  to  the 
middle  point  of  the  back  of  a  chair  will  probably  cause  the  chair 
to  overturn.  A  similar  force  applied  to  one  foot  will  drag  it  along 
the  ground  and  also  cause  it  to  turn  about  a  vertical  axis. 

The  position  occupied  by  the  particle  to  which  a  force  is  applied 
is  called  the  foint  of  application  of  the  force.  The  line  drawn 
through  this  point  in  the  direction  of  the  force  is  called  the  line 
of  action  of  the  force. 

Clearly,  in  order  to  have  full  data  as  to  the  action  of  a  force, 
we  must  know 

(a)  its  magnitude; 

(h)  its  point  of  application; 

ic)  its  line  of  action. 

\^  Moments 

46.  Definition.  The  moment  of  a  force  ahout  a  line  at  right 
angles  to  the  line  of  action  of  the  force  is  defined  to  he  the  product 
of  the  force  and  of  the  shortest  distance  hetween  the  two  lines. 

This  moment,  as  we  shall  soon  find,  measures  the  tendency  to  turn  aronnd 
the  line  about  which  the  moment  is  measured  ;  e.g.  if  the  arm  of  a  balance 
is  of  length  /,  a  weight  in  at  its  end  has  a  moment  Iw  about  the  pivot  of  the 
balance,  and  we  shall  find  that  this  measures  the  tendency  of  the  arm  to  turn. 

Definition.  The  moment  of  a  force  about  a  line  L  which  is  not 
at  right  angles  to  the  force  is  defined  to  be  the  saine  as  the  moment 
about  Lj)f  the  component  of  the  force  in  a  plccne  perpendicular  to  L. 


MOMENTS  61 

Resolving  the  force  into  two  comjionents,  one  parallel  to  L  and  one 
perpendicular  to  Z,  it  is  clear  that  the  former  will  not  give  any  tendency 
to  turn  about  L,  so  that  the  whole  tendency  to  turn  conies  from  the  second 
component. 

The  two  definitions  which  have  now  been  given  suffice  to  deter- 
mine the  moment  of  any  force  F  about  any  line  L.  It  may  be 
noticed  that  the  moment  vanishes 

(a)  if  the  line  of  action  of  F  is  parallel  to  L ; 

(b)  if  the  line  of  action  of  F  intersects  L. 

Obviously,  in  either  of  these  cases,  the  tendency  to  turn  about  L  is  zero. 

47.  Let  the  line  L  be  at  right  angles  to  the  plane  of  the  paper, 
and  intersect  it  in  the  point  M.    Let  PA  be  the  line  of  action  of 
a  force  F  in  the  plane  of  the  paper,  acting  on  a  particle  at  A,  and 
let  MN  be  the  perpendicular  from  M  on 
to  FA.    Then,  by  definition,  the  moment 
of  the  force  F  about  L  \&  F  x  MN. 

Let  the  angle  PAS  be  drawn  equal 
to  the  angle  NMA,  say  equal  to  6,  so 
that  AS  is  perpendicular  to  MA.  Then 
the  moment  of  the  force  F  about  L 

=  Fx  MN 

=  FxA3Ico&d  p^^  32 

=  A3I X  Fcosd 

=  AM  X  resolved  part  of  F  along  SA. 

Instead  of  F  being  the  actual  force  acting  at  A,  suppose  that  F 
is  the  resolved  part,  in  the  plane  perpendicular  to  the  line  L,  of 
some  other  force  B.  Then  the  moment  of  E  about  L  is,  by  defini- 
tion, the  same  as  the  moment  of  F,  and  the  resolved  part  of  R 
along  AS  =  F cos  0.  Hence  what  has  just  been  proved  may  be 
put  in  the  form 

moment  about  L  of  any  force  B  acting  at  A 

=  AM  X  resolved  part  of  B  along  SA, 

and  SA  is  now  determined  as  the  direction  wliich  is  perpendicular 
to  L,  and^lso  to  A3I,  the  perpendicular  from  A  on  to  L. 


62  STATICS  OF   SYSTEMS   OF  PARTICLES 

Thus  we  have  a  new  definition  of  a  moment,  which  is  exactly 
equivalent  to  that  previously  given,  namely: 

The  moment  about  a  line  L  of  a,  force  R  acting  at  A  is  equal 
to  AM,  the  perpendicular  from  A  on  to  L,  multiplied  hy  the  com- 
ponent  of  Ji  in  a  direction  perpendicular  to  AMand''toIj.  ' 

48.  From  this  conception  of  a  moment  we  have  at  once  the 
theorem : 

Tlie  sum  of  the  moments  ahout  any  line  L  of  any  number  of 
forces  acting  at  a  point  A  is  equal  to  the  moment  of  their  resultant 
about  L. 

For,  let  ^1,  i^2'  ■  ■  ■  ^^  the  forces,  and  B  their  resultant.  Let  AM, 
as  in  §  47,  be  the  perpendicular  from  A  on  to  L,  and  let  ^*S^  be  a 
direction  perpendicular  to  AM  and  to  L.    The  theorem  to  be  proved 

is  that 

AM  X  component  of  ii\  along  AS 

4-  AM X  component  of  li^  along  AS  +  ■■■ 

=  AM  X  component  of  B  along  AS. 

On  dividing  through  by  AM  the  theorem  to  be  proved  is  seen 
to  be  simply  that  the  component  of  R  along  AS  is  equal  to  the 
sum  of  the  components  of  R^,  R^,  ■■■  along  AS,  which  is  known 
to  be  true. 

We  can  now  see  more  clearly  how  it  is  that  the  moment  of  a 
force,  defined  as  we  have  defined  it,  gives  a  measure  of  the  tendency 
to  turn.  In  fig.  32  we  are  taking  moments  about  a  line  L  which 
is  at  right  angles  to  the  plane  of  the  paper  and  meets  this  plane 
at  M.  The  force  whose  moment  is  being  considered  is  a  force  R 
acting  at  the  point  A.  At  A  we  have  three  directions  mutually  at 
right  angles,  namely 

AS,  AM,  and  the  direction  of  a  line  through  A  parallel  to  L. 

The  moment  of  R  abovit  L  has  been  defined  to  be 

A3I  X  component  of  R  along  AS. 


MOMENTS  63 

Now  the  component  of  E  along  AM  is  a  force  of  wiiich  the  Ime 
of  action  intersects  L,  and  so  can  produce  no  tendency  to  turn  a 
body  about  L,  while  the  component  of  li  along  the  line  through  A 
parallel  to  L  can  again  produce  no  tendency  to  turn  about  L.  Thus 
B  can  be  resolved  into  three  components,  of  which  only  the  first, 
the  component  along  AS,  tends  to  set  up  rotation  about  L.  We 
have  defined  the  moment  of  the  whole  force  E  in  such  a  way  that 
it  becomes  identical  with  the  moment  of  that  one  of  its  components 
which  tends  to  set  up  rotation. 

It  will  be  noticed  that  a  moment  has  sign  as  well  as  magnitude. 
In  moving  along  the  line  of  action  of  a  force  E,  we  may  turn  in 
either  one  direction  or  the  other  about  a  line  L.  We  agree  that 
when  the  turning  is  in  one  direction  the  moment  of  E  about  L  is 
to  be  regarded  as  positive ;  when  the  turning  is  in  the  other 
direction  the  moment  is  taken  to  be  negative. 

49.  If  a  particle  is  in  equilibrium  under  the  action  of  any  num- 
ber of  forces,  the  resultant  of  all  these  forces  must  be  nil.  The 
sum  of  the  moments  of  the  separate  forces,  taken  about  any 
line  whatever,  is  equal  to  the  moment  of  the  resultant  and  is 
therefore  nil. 

Hence  we  have  the  result : 

When  a  j^a/rtide  is  in  equilibrium  under  the  action  of  any  forces, 
the  su7n  of  the  moments  of  these  forces  about  any  line  wKafever 
must  vanish. 


System  of  Particles  in  Equilibrium 

50.  Consider  a  system  of  particles  siqtposed  to  be  m  equilibrium 
under  the  action  of  any  number  of  forces.  As  we  have  seen,  the 
forces  actmg  on  any  single  particle  will  be  of  two  kinds: 

(a)  external  forces,  forces  applied  to  the  particle  from  outside, 
as  for  instance  the  weight  of  the  particle; 

(h)  internal  forces,  forces  of  interaction  between  the  particle  and 
the  remaining  particles  of  the  system. 


64  STATICS  OF  SYSTEMS  OF  PARTICLES 

Now  if  the  whole  system  of  particles  is  in  equilibrium,  it  follows 
that  each  particle  separately  must  be  in  equilibrium.  It  follows 
from  §  33,  that 

(a)  the  sum  of  the  components  in  any  direction,  of  all  the  forces 
acting  on  any  single  particle,  must  vanish; 

and  from  the  theorem  just  proved  in  §  48,  that 

(b)  the  sum  of  the  moments  about  any  line,  of  all  the  forces 
acting  on  any  single  particle,  must  vanish. 

If,  however,  the  sum  of  the  components  of  the  forces  acting  on 
each  particle  vanishes,  it  follows  by  addition  that  the  sum  of  the 
components  of  all  the  forces  acting  on  all  the  particles  must  vanish. 
The  sum  of  the  components  of  the  internal  forces,  however,  van- 
ishes by  itself,  for  the  internal  forces  consist  of  pairs  of  actions 
and  reactions,  and  the  two  components  in  any  direction  of  such  a 
pair  of  forces  are  equal  and  opposite. 

Since  the  total  sum  vanishes,  and  the  sum  of  the  components 
of  internal  forces  vanishes,  it  follows  that  the  sum  of  the  com- 
ponents of  external  forces  vanishes. 

A  similar  proposition  is  true  of  the  moments  of  the  external 
forces.  The  sum  of  the  moments  abovit  any  line  L  of  all  the  internal 
forces  is  nil,  for  the  moments  of  an  action  and  reaction  are  ec^ual 
and  opposite.  The  sum  of  the  moments  of  all  the  forces,  internal 
and  external,  is  zero,  for  each  sum  of  the  moments  of  the  forces 
acting  on  each  particle  is  zero  separately.  Thus  the  sum  of  the 
moments  of  the  external  forces  is  zero. 

Thus  we  have  proved  the  following  theorems : 

When  a  system  of  particles  is  in  equilibrium  under  the  action 
of  any  system  of  external  forces, 

(«)  the  sum  of  the  components  of  all  these  forces  in  any  direction 
is  zero  ; 

(I))   the  sum  of  the  moments  of  all  these  forces  about  any  line  is  zero. 

Speaking  loosely,  we  may  say  that  these  theorems  express  that 
there  is  no  tendency  to  advance  in  any  direction  or  to  turn  about 
any  line. 


ILLUSTRATIVE   EXAMPLE 


65 


ILLUSTRATIVE  EXAMPLE 


Wheel  and  axle.  The  apparatus  known  as  the  "wheel  and  axle"  consists  of 
a  circular  axle  free  to  turn  about  its  central  axis,  to  which  a  circular  wheel 
is  rigidly  attached,  so  that  its  center  is  on  the  center  of  the  axle.  A  rope  or 
string  is  wound  round  the  axle,  and  has  a  weight  attached  to  its  end.  A  second 
rope  or  string  is  wound  round  the  circumference  of  the  circle  in  the  opposite 
direction,  and  this  again  has  a  weight  attached  to  its  end.  By  a  suitable  choice 
of  the  ratio  of  these  two  weights,  the  apparatus  may  be  balanced  so  that  there 
is  no  tendency  for  it  to  turn  about  its  axis. 

Let  us  consider  the  equilibrium  of  the  system  consisting  of  the  wheel  and 
axle  and  of  those  parts  of  the  strings  or  ropes  which  are  wound  round  them. 
To  simplify  the  problem,  let  us  disregard  altogether  the  weight  of  the  system. 
Then  the  externally  applied  forces  are 

(a)  the  tension  of  the  rope  wound  round 
the  wheel ; 

(6)  the  tension  of  the  rope  wound  round 
the  axle ; 

(c)  the  action  of  the  supports  which  keep 
the  wheel  and  axle  from  falling. 

Let  the  weights  be  denoted  by  P  and  Q,  so 
that  these  are  also  the  tensions  of  the  strings, 
and  let  the  radii  of  the  wheel  and  axle  be  a, 
b  respectively.  Let  us  express  mathematically 
that  the  sum  of  the  moments  of  the  externally 
applied  forces  about  the  axis  is  nil. 

The  moment  of  the  tension  of  the  string 
on  the  wheel  is  Pa,  for  P  is  the  amount  of 
the  tension,  which  acts  at  right  angles  to  the 
axis,  and  a  is  the  shortest  distance  from  the 
axis  to  the  line  of  action  of  this  tension. 

Similarly  the  moment  of  force  (b)  is  —  Qb, 
the  negative  sign  being  taken  because  this  tends  to  turn  the  system  in   the 
direction  opposite  to  that  in  which  the  first  tension  tends  to  turn  it. 

If  we  imagine  the  system  to  be  supported  by  forces  acting  on  the  axis  itself, 
the  moment  of  forces  (c)  vanishes,  for  the  lines  of  action  of  these  forces  intersect 
the  line  about  which  we  are  taking  moments.    Thus  the  required  equation  is 

Pa-  Qb  =  0. 

This  equation  simply  expresses  that 

[tendency  of  P  to  turn  system]  —  [tendency  of  Q  to  turn  system]  =0. 

Thus  when  the  system  is  balanced  so  as  to  remain  at  rest  we  must  have 

P:Q  =  b:a, 
so  that  the  weights  must  be  inversely  as  the  radii.    Practical  examples  of  the 
principle  of  the  wheel  and  axle  are  supplied  by  the  windlass  and  capstan. 


Fig.  33 


66  STATICS  OF   SYSTEMS  OF  PARTICLES 


EXAMPLES 

'j^l.  Eight  sailors,  each  pressing  on  the  arm  of  a  capstan  with  a  horizontal  force 
of  100  lbs.,  at  a  distance  of  8  feet  from  its  center,  can  just  raise  the  anchor. 
The  radius  of  the  axle  of  the  capstan  is  12  inches.  Find  the  pull  on  the  cable 
which  raises  the  anchor. 

^2.  In  the  apparatus  of  fig.  33  the  weight  P  is  disconnected,  and  the  free  end 
of  the  string  is  tied  to  the  same  point  on  Q  as  the  other  string.  Show  that  in 
equilibrium  this  point  is  vertically  below  the  axis. 

^/3.  A  wheel  is  free  to  turn  about  a  horizontal  axis,  and  has  fastened  to  it 
two  strings  which  are  wound  round  its  circumference  in  opposite  directions. 
The  other  ends  are  both  tied  to  a  small  ring  from  which  a  weight  is  suspended. 
Show  that  when  the  system  is  at  rest  the  two  strings  will  make  equal  angles 
wixh  the  vertical. 

•/4.  A  man  finds  that  he  can  just  move  a  lock  gate  against  the  pressure  of  the 
water,  by  pressing  with  a  horizontal  force  of  150  lbs.  at  a  distance  of  8  feet 
from  the  pivot.  What  force  must  he  exert  if  he  presses  at  a  distance  of  9  feet 
from  the  pivot  ? 

5.  A  wheel  capable  of  turning  freely  about  a  horizontal  axis,  has  a  weight 
of  2  pounds  fixed  to  the  end  of  a  spoke  which  makes  an  angle  of  60°  with  the 
horizontal.  AVhat  weight  must  be  attached  to  the  end  of  a  horizontal  spoke  to 
prevent  motion  taking  jalace  ? 

6.  A  drawbridge  is  raised  by  a  chain  attached  to  the  end  farthest  removed 
from  the  hinges.  When  the  bridge  is  at  rest  in  a  horizontal  position,  the  chain 
makes  an  angle  of  60°  with  the  bridge,  and  the  pull  on  the  chain  necessary  to 
move  the  bridge  is  equal  to  the  weight  of  three  tons.  Find  what  additional 
pull  is  required  in  the  chain  when  a  weight  of  one  ton  is  placed  at  the  middle 
point  of  the  bridge. 


Forces  ix  one  Plane 

51.  The  simplest  problems  in  statics  are  always  those  in  which 
all  the  forces  have  their  lines  of  action  in  one  plane.  In  such  a 
problem  it  is  obviously  most  convenient  to  take  moments  about 
a  Ime  perpendicular  to  the  plane  in  which  the  forces  act.  Let  any 
such  Ime  intersect  the  plane  in  a  point  P.  Each  force  is  entirely 
perpendicular  to  the  line  about  which  moments  are  taken,  so  that 
the  moment  is  equal  to  the  product  of  the  force  and  the  shortest 
distance  of  the  line  of  action  of  the  force  from  P. 

Taking  moments  about  an  axis  which  intersects  the  plane  of 
the  forces  at  right  angles  in  a  pomt  P  is  often  spoken  of  as  taking 


CO-PLANAR  FORCES  67 

onovients  about  tJie  point  P,  and  the  perpendicular  from  P  to  the 
line  of  action  of  a  force  is  spoken  of  as  the  arm  of  the  moment 
of  this  force. 

52.  Theorem.  When  three  forces,  acting  in  a  plane,  keep  a  hocly 
or  system  of  bodies  in  eqitilibriuin,  these  three  forces  must  tneet  in 
a  point. 

For  let  P,  Q,  R  be  the  forces,  and  let  P,  Q  intersect  in  the 
point  A.  Then  the  sum  of  the  moments  of  P,  Q,  and  R  about  A 
must  vanish,  and  those  of  P  and  Q  are  already  known  to  vanish. 
Thus  the  moment  of  R  about  A  must  vanish,  —  that  is,  R  must 
pass  through  the  point  A,  or,  what  is  the  same  thing,  the  three 
forces  must  intersect  in  a  single  point. 

An  application  of  this  principle  is  often  sufficient  in  itself  for 
the  solution  of  statical  problems  in  which  the  applied  forces  can 
be  reduced  to  three. 

ILLUSTRATIVE  EXAMPLES 

1.  The  seesaw.  Two  persons  of  weights  TFi,  Wn  stand  on  a  plank  ichich  rests 
on  a  rough  support  about  which  it  is  free  to  turn.  Neglecting  the  weight  of  the 
plank,  find  how  the  persons  must  place  themselves  in  order  that  the  plank  may 
balance. 

The  forces  may  be  supposed  all  to  act  in  one  plane,  namely  the  vertical 
plane  through  the  central  line  of  the 
plank.    The  forces  are 

(a)  the  weight  Wi  of  the  person  at 
one  end  ;  ^ 


(b)  the  weight  W-z  of  the  person  at         ]\\  /\  " .: 
the  other  end  ;                                                                         /      \ 

(c)  the  reaction  between  the  plank  „ 

,    .  '■  Fig.  34 

and  Its  support. 

Let  a,  b  be  the  distances  of  the  persons  from  the  support ;  then,  on  taking 
moments  about  the  point  of  support,  we  have 

Wia  -W.2b  =  0. 

Thus  the  two  persons  should  stand  at  distances  from  the  support  which  are 
inversely  proportional  to  their  weights. 

Notice  that  in  this  problem  the  system  is  acted  on  by  three  forces,  which  meet  in 
a  point,  the  point  being  at  infinity. 


68 


STATICS   OF   SYSTEMS   OF  PARTICLES 


Fig. 


2.  The  nutcracker.  It  is  found  that  a  weight  of  100  pounds  placed  on  top  of  a 
nut  will  just  crack  it.  How  much  force  must  be  applied  at  the  ends  of  the  arms  of 
a  nutcracker  6  inches  long  to  crack  the  nut  when  it  is  placed  h  inch  from  the  hinge  ? 

Let  a  force  F  applied  at  the  extreme  end  of  each  arm  be  supposed  just  suffi- 
cient to  crack  the  nut.  Then  when  a  force  F  is  applied  at  the  end  of  the  arm,  the 
pressure  between  the  nut  and  the  arm  must  be  the  weight  of  100  pounds.    Thus 

the  forces  acting  from  outside  on  either  arm 
of  the  nutcracker  will  consist  of 

(a)  tlie  force  F  applied  at  tlie  end  of  the  arm  ; 
(6)  the  pressure  of  100  pounds  weight  exerted 
by  tlae  nut  on  the  arm  at  a  distance  of  ^  inch 
from  the  liinge  ; 

(c)  tlie  reaction  at  the  hinge. 
The  weight  of  the  nutcracker  is  here  supposed 
to  be  negligible. 

Taking  moments  about  tlie  liinge,  we  obtain 
C  X  F  =  I  X  100  pounds  weight, 
so  that  F  =  8i  pounds  weight. 

Note.  When,  as  here,  an  unknown  force  neither  enters  in  the  data  nor  is  required 
in  the  answer,  we  can  always  obtain  equations  in  which  the  force  does  not  occur,  by 
taking  moments  about  a  point  in  its  line  of  action.  So  again,  if  two  such  forces  occur, 
we  can  obtain  an  equation  into  which  neither  force  enters,  by  taking  moments  about 
the  point  of  intersection  of  their  two  lines  of  action. 

3.  'A  ladder  stands  on  a  rough  horizontal  plane,  leaning  against  a  rough  ver- 
tical wall,  the  contacts  at  the  two  ends  of  the 
ladder  being  equally  rough.  Find  how  far  a 
man  can  ascend  the  ladder  without  its  slipping, 
it  being  supposed  that  the  weight  of  the  ladder 
may  be  neglected. 

The  forces  acting  on  the  system  compo.sed 
of  the  man  and  ladder  are  three  in  number : 
(a)  the  reaction  with  the  horizontal  plane  ; 
(6)  the  reaction  with  the  vertical  wall ; 
(c)  the  weight  of  the  man. 

The.se  forces  are  all  in  one  plane ;  hence, 
by  the  theorem  of  §  52,  their  lines  of  action 
must  meet  in  a  point. 

In  the  figure  let  AB  be  the  ladder,  C 
the  position  of  the  man,  and  P  the  point  in 
which  the  three  forces  meet,  so  that  PC  is 
vertical,  and  AP,  BP  are  the  lines  of  action 
of  the  reactions  at  A,  B.  When  slipping  is  just  about  to  begin,  each  of  these 
reactions  must  make  with  the  normal  an  angle  equal  to  the  angle  of  friction. 


Fig.  .36 


ILLUSTRATIVE   EXAMPLES 


69 


Let  e  be  this  angle  of  friction,  and  let  a  be  the  inclination  of  ladder  to  the 
horizontal.    Then,  from  the  geometry  of  the  triangle  ACP,  we  have 

AC AP 

sine 


and  since  APB  is  a  right  angle. 


Thus 


AP  =  AB  cos  / 
A  C  =  AP  sin  e  sec  a  =  AB  sin  e  sin  (e  +  a)  sec  a. 


Thus  slipping  will  begin  as  soon  as  the  man  has  climbed  a  height  equal  to 
sin  e  sin  (e  +  a)  sec  a  times  thecwhole  height. 

The  condition  that  the  man  can  reach  the  top  without  slipping  is  that 
sin  e  sin  (e  +  a)  sec  a  shall  be  greater  than  unity,  or  that 

sin  e  sin  (e  +  a)  >  cos  [(e  +  a)  —  e] 

>  sin  e  sin  (e  +  a)  +  cos  e  cos  (e  +  a). 

Thus  for  the  condition  to  be  satisfied  cos  e  cos  (e  +  a)  must  be  negative  ;  i.e, 
e+  a  must  be  greater  than  90°.  Thus  the  angle  between  the  ladder  and  the 
vertical  must  be  less  than  the  angle  of  friction.  This  is  also  clear  from  the 
figure,  for  when  the  man  reaches  B,  two  of  the  forces,  namely  the  reaction  at  B 
and  the  weight  of  the  man,  both  pass  through  B,  so  that  the  third  force  must 
also  pass  through  B;  i.e.  the  reaction  at  A  must  have  AB  for  its  line  of  action, 
and  if  the  ladder  just  slips  here,  the  angle  between  AB  and  the  vertical 
must  be  e. 

4.  If,  in  the  last  problem,  the  man  has 
ascended  to  some  point  C  without  the  ladder 
slipping,  what  are  the  reactions  at  A  and  B  f 

Here  it  is  not  known  what  angles  the 
reactions  make  with  the  normals:  all  that 
is  known  is  that  these  angles  are  less  than 
the  angle  of  friction. 

Let  us  resolve  the  reaction  at  A  into  two 
components  Nx,  Fi,  and  that  at  B  into  two 
components  N2,  F^,  these  being  horizontal 
and  vertical  as  in  the  figure.  Then  the 
forces  acting  on  the  .system  composed  of 
the  man  and  ladder  are  the  five  forces 
Ni,  Fi,  No_,  F.2,  and  W. 

Resolving  vertically. 
Resolving  horizontally. 
Taking  moments  about  A, 

W ■  AC  CO?,  a  -  Fn. 


forces                    ■"■        jr 

Fig.  37 

TF-.Yi  -F2  =  0. 

(a) 

Fi  -  N.  =  0. 

(&) 

AB  cos  a  —  N-i  ■  AB  sin  a:  =  0. 

(c) 

70 


STATICS  OF   SYSTEMS   OF  PARTICLES 


There  are  four  quantities  wliich  it  is  required  to  find.  So  far  we  liave  obtained 
only  tliree  equations.  We  can,  of  course,  obtain  otlier  equations  by  resolving 
in  other  directions  and  by  taking  moments  about  other  points,  but  it  will  be 
found  that  the  equations  so  obtained  will  not  be  new  equations,  but  simply 
equations  of  which  the  truth  is  already  implied  in  the  equations  already 
obtained.  Thus  we  cannot,  by  resolving  and  taking  moments,  obtain  more  than 
three  independent  equations,  and  these  do  not  suffice  to  determine  the  four 
unknown  quantities. 

Here  we  have  illustrated  a  problem  which  cannot  be  solved  by  the  methods 
explained  in  this  chapter,  and  which  requires  for  its  solution  a  consideration 
of  the  systems  of  forces  set  up  between  the  separate  particles  of  the  bodies  acted 
upon.  It  is  important  tliat  the  student  should  realize  that  such  problems  exist, 
although  he  may  not  yet  be  able  to  solve  them. 

5.  Force  required  to  drag  a  car  along.  To  simplify  the  problem  as  faraS- 
possible,  let  us  suppose  that  the  car  is  mounted  on  four  equal  wheels,  each  of 
radius  a  and  revolving  round  an  axle  of  radius  b,  the  coefficient  of  friction 
between  wheel  and  axle  being  the  same  for  each  wheel.  Let  us  suppose  that  a 
force  P  applied  horizontally  is  found  to  be  just  sufficient  to  start  the  car  into 
motion. 


,> 


^ 


Fig.  38 


Let  us  consider  first  the  equilibrium  of  the  whole  car.  We  are  most  con- 
veniently able  to  enumerate  the  forces  acting  on  any  system  by  taking  a  tour, 
in  imagination,  over  the  whole  surface  of  a  cover  made  just  to  fit  the  system, 
and  noting  the  forces  which  act  across  this  cover  at  its  different  points.  These, 
together  with  the  weight  of  the  whole  system,  will  give  the  whole  system  of 
forces.    The  forces  acting  on  the  car  are  in  this  way  found  to  be 

(a)  its  weight,  say  W; 

(6)  the  horizontal  applied  force  P  ; 

(c)  the  reactions  between  the  wheels  and  the  ground.  Let  us  resolve  each 
reaction  into  a  vertical  component  R  and  a  horizontal  component  F;  let  us 
denote  the  components  of  the  reaction  between  the  first  wheel  and  the  ground 
by  Pi,  Pi ;  let  the  corresponding  quantities  for  the  second  wheel  be  P2,  P2 ;  and 
so  on. 


ILLUSTRATIVE  EXAMPLES 


71 


(As  regards  the  f rictional  force  acting  between  the  wheel  and  the  ground,  we 
notice  that  although  motion  is  about  to  take  place,  this  motion  is  not  one  of 
slipping  between  the  wheel  and  the  ground,  so  that  the  ratio  of  i?"  to  ii  for  any 
wheel  is  not  the  coefficient  of  friction  between  the  wheel  and  the  ground.) 

The  forces  just  enumerated  hold  the  car  in  equilibrium.  Thus  the  sum  of 
their  components  in  any  direction  must  vani.sh  and  the  sum  of  their  moments 
about  any  line  must  vanish.    Resolving  horizontally  and  vertically,  we  obtain 

P  =  Fi  +  F2  +  F3  +  Fi.  (a) 

W  =  Ri+Rz  +  R3  +  Bi.  (b) 

There  is  nothing  to  be  gained  by  taking  moments  about  any  line  ;  as  we  do  not 
know  the  line  of  action  of  P,  we  cannot 
know  its  moment. 

Next  let  us  consider  the  equilibrium  of 
a  single  wheel.  The  wheel  touches  the 
ground  and  also  touches  the  axle  at  some 
point  C.  (We  may  think  of  the  axle  as  a 
circle  of  radius  veiy  slightly  less  than 
that  of  the  inside  of  the  hub  of  the 
wheel.)  The  forces  acting  on  the  wheel 
are  accordingly 

(a)  its  reaction  with  the  ground  ; 

{b)  its  reaction  with  the  axle  ; 

(c)  its  weight,  which  we  shall  neglect 
as  being  insignificant  in  comparison  with 
that  of  the  car. 

Neglecting  the  third  force,  the  two  fonner  forces  must  be  equal  and  opposite. 
The  line  of  action  of  each  is  accordingly  the  line  joining  B,  the  point  of  contact 
with  the  ground,  to  C  the  point  of  contact  with  the  axle.  Since  slipping  is  just 
about  to  take  place  at  C,  the  reaction  at  C  will  make  an  angle  e,  equal  to  the 
angle  of  friction,  with  A  C  the  normal  at  C.    Thus  the  angle  BCA  is  equal  to  e. 

In  the  triangle  A  CB  we  have  AC  =  b,  AB  =  a,  and  the  angle  ACB  —  e. 

a    _        ^ 
sine      sitiABC 


Thus 


Since,  however,  a  force  along  BC  has  comi)onents  Ei  and  Fi  along  and  per- 
pendicular to  AB,  we  have 

tan^JSC  =  — • 


Thus 


^1 


Ri 
=  tan  ABC 


?.mABC 


Vl  -  sin2J.BC 
b  sin  e 

Va'-  —  b'^  sin^  e 


72  STATICS  OF   SYSTEMS   OF  PARTICLES 

Kovv  e,  a,  and  b  are  supposed  to  be  the  same  for  each  wheel,  so  that 
6  sin  6  Fi       F2      Fs      F4 


Va^  —  62  sin^  e     -^1      ^2      ^3      ^4 
^Fi  +  F2  +  F3  +  F4: 
Ri  +  R2  +  lis  +  Ri 
_  P 

by  equations  (a)  and  (6). 

r^,  -r^  Wb  sin  e  ,  , 

Thus  P  =  — ,  .  (c) 

V a2  —  62  sin2  e 

giving  the  horizontal  pull  required. 

The  value  of  6,  the  radius  of  the  axle,  will  generally  be  small  in  comparison 
with  a,  the  radius  of  the  wheel.  Thus,  without  serious  error,  we  may  neglect 
52  sin2  e  in  comparison  with  a2,  and  replace  the  denominator  in  equation  (c)  by  a. 
The  equation  now  becomes 

_  Wb  sin  e 

~        a 

By  making  b/a  very  small,  we  see  that  the  car  can  be  made  to  run  very 
smoothly.  We  notice  also  that  even  if  there  is  so  much  friction  between  wheel 
and  axle  that  the  coefficient  of  friction  may  be  regarded  as  infinite,  we  have 
sine  =  1,  and  hence 

„      Wb 


so  that  the  force  required  to  drag  the  car  along  will  still  be  small  compared 
with  that  required  to  drag  the  same  weight  over  a  fairly  smooth  surface. 

This  analysis  has  assumed  that  the  wheels  may  be  supposed  to  touch  the 
ground  only  at  their  lowest  point.  It  applies  pretty  accurately  to  the  case  of 
steel  wheels  rolling  on  steel  rails,  but  does  not  apply  to  the  problem  of  an 
ordinary  road  carriage  moving  over  a  soft  road,  where  the  wheels  are  embedded 
to  a  small  extent  in  the  road.  In  fact,  if  the  analysis  just  given  took  account  of 
all  the  facts  of  the  case,  it  is  clear  that  the  force  required  to  haul  a  car  would 
be  independent  of  the  state  of  the  road. 


EXAMPLES 

w_  1.  A  weight  of  250  pounds  is  suspended  from  a  light  rod  which  is  placed 
/  over  the  shoulders  of  two  men  and  carried  in  a  horizontal  position.  If  the  men 
walk  10  feet  apart  and  the  weight  is  4  feet  from  the  nearer  of  them,  find  the 
weight  borne  by  each, 
■y^  2.  A  weight  is  suspended  from  a  light  rod  which  passes  over  two  fixed  sup- 
ports 6  feet  apart.  On  moving  the  weight  (i  inches  nearer  to  one  support,  the 
pressure  on  that  support  is  increased  by  10  pounds.  What  is  the  amount  of  the 
■weight  ? 


EXAMPLES  73 

y  3.  A  balance  has  two  pans,  each  of  weight  8  ounces,  suspended  from  a  beam, 
each  at  distance  7  inches  from  the  pivot.  A  dishonest  tradesman  moves  one 
pan  half  an  inch  nearer  to  the  pivot,  adding  weight  to  the  farther  pan  in  order 
that  the  two  pans  may  still  balance.  Find  how  much  weight  he  must  add,  and 
by  hc^  much  his  profits  will  be  increased  by  his  dishonesty. 

y^.  A  balance  has  a  weight  of  20  ounces  suspended  from  one  end  of  the  beam. 
A  string  is  tied  to  the  other  end  of  the  beam  and  at  equal  distance  from  the 
pivot,  and  this  string  makes  an  angle  of  45°  with  the  horizontal.  With  what  force 
must  it  be  pulled  to  maintain  the  beam  of  the  balance  in  a  horizontal  position  ? 

5.  A  crowbar  8  feet  long  is  to  be  vised  to  move  a  body,  it  being  required  to 
apply  a  force  of  500  pounds  weight  vertically  upwards  to  this  body  to  move  it. 
How  near  to  the  end  of  the  crowbar  must  the  fulcrum  be  placed  in  order  that  a 
man  of  140  pounds  weight  may  be  able  to  apply  the  required  force  by  pressing 
on  the  other  end  of  the  crowbar  ? 

6.  A  table  of  negligible  weight  has  any  number  of  legs.  A  heavy  particle  is 
placed  on  the  table.  Show  that  the  table  will  tilt  over  if  the  vertical  through 
the  particle  meets  the  floor  under  the  table,  in  a  point  outside  the  polygon 
formed  by  joining  the  points  of  contact  of  the  feet  with  the  floor. 

7.  A  table  of  negligible  weight  has  three  legs,  the  feet  forming  an  equilateral 
triangle.  A  heavy  particle  is  placed  on  the  table  in  a  position  such  that  the  table 
does  not  tilt  over.    Find  the  proportion  of  weight  which  is  carried  by  each  foot. 

8.  A  card  is  suspended  in  a  horizontal  position  by  three  equal  inextensible 
strings  fastened  to  three  points  A,  B,  C  in  the  card  which  form  an  equilateral 
triangle,  and  also  to  a  point  P  above  the  card.  A  weight  is  placed  on  the  card 
at  any  point  Q  inside  the  triangle  ABC.    Find  the  tensions  of  the  strings. 

9.  A  card  is  suspended  by  four  equal  inextensible  strings  which  pass  through 
the  four  points  A,  B,  C,  D  of  a  square  in  the  card  and  are  tied  to  four  points 
A',  B\  C,  D'  at  equal  heights  h  vertically  above  the  points  A,  B,  C,  D.  A 
weight  is  placed  on  the  card  at  any  point  P  inside  the  square  ABCD.  Show 
that  the  tensions  in  the  strings  cannot  be  determined  without  discussing  the 
internal  stresses  in  these  strings. 

10.  If  in  the  last  question  the  internal  stresses  in  the  strings  stretch  the 
strings  very  slightly,  so  that  Hooke's  law  is  obeyed,  show  that  the  tensions  can 
be  found,  and  find  them. 

^11.  A  seesaw  rolls  on  a  rough  circular  log  of  radius  a,  fixed  horizontally. 
Two  persons  stand  at  distances  b,  c  from  the  middle  point  of  the  seesaw,  tlieir 
weights  being  such  that  the  seesaw  is  just  balanced  horizontally  with  the 
middle  point  resting  on  the  log.  The  first  person  moves  a  distance  d  towards 
the  center  of  the  log.  Through  what  angle  will  the  seesaw  turn  ?  How  far 
can  the  person  advance  before  the  seesaw  slips  off  the  log  altogether? 
y  12.  A  pair  of  wheels  of  radius  a  are  connected  by  an  axle  of  radius  b,  and  run 
on  horizontal  rails.  A  string  is  wound  round  the  axle  and  the  end  leaves  the 
axle  making  an  angle  6  with  the  horizontal.  If  this  string  is  pulled,  show  that 
the  wheels  will  run  toward  or  away  from  the  person  pulling  according  as 
cos  e  is  greater  or  less  than  b/a.    What  happens  when  cos  d  =  b/a '? 


t4  STATICS   OF   SYSTEMS   OF  PARTICLES 

13.  If  the  weight  of  the  wheels  and  axles  of  a  car  is  w,  and  if  this  may  not 
be  neglected  in  comparison  with  W,  the  total  weight  of  the  car,  show  that 
equation  (c)  of  example  5,  p.  72,  must  be  replaced  by 

_  (W  —  w)b sin  e 

V  a^  —  b-  sin^  e 

14.  A  locomotive  of  weight  134  tons  rests  on  a  bogie,  of  which  the  wheels  and 
axles  weigh  4  tons,  and  two  pairs  of  driving  wheels,  of  which  the  wheels  and 
axles  weigh  10  tons.  The  weight  taken  on  the  axles  of  the  bogie  is  40  tons, 
that  taken  on  the  axles  of  the  driving  wheels  being  80  tons.  The  diameters  of 
the  wheels  are  2  feet  10  inches  and  7  feet  1  inch  respectively.  Each  axle,  where 
it  passes  through  the  axle  box,  is  of  radius  2^  inches,  and  the  coefficient  of  fric- 
tion is  l-    Find  the  horizontal  force  necessary  to  move  the  engine. 

15.  In  question  14  the  rails  are  greased  so  that  the  coefficient  of  friction 
between  them  and  the  wheels  is  less  than  yi^ ;  show  that  the  engine  cannot 
be  started  without  the  wheels  skidding  on  the  greased  rails,  and  explain  the 
dynamical  processes  by  which  the  engine  is  set  in  motion  in  this  case. 


StRINCxS 

53.  Strings,  ropes,  and  chains  frequently  form  part  of  the  systems 
of  bodies  witli  which  statical  problems  are  concerned,  so  that  it  is 
important  to  discuss  the  equilibrium  of  a  string  (or  rope  or  cham). 
The  first  problem  we  shall  consider  is  that  of  a  string  stretched 
over  a  surface, —  as  for  example  a  pulley  wheel, —  it  l)eing  supposed 
that  the  weight  of  the  string  may  be  neglected,  and  that  the  con- 
tact between  the  string  and  the  surface  is  equally  rough  at  all 
points.    It  will  also  be  supposed  that  the  string  is  all  in  one  plane. 

Let  F,  Q  be  two  adjacent  points  of  the  string  so  near  together 
that  the  portion  PQ  oi  the  string  may  be  treated  as  a  particle. 

The  forces  acting  on  this  particle  will  be 

(a)  Tp,  the  tension  at  P,  acting  along  the  tangent  to  the  string 
at  P; 

(1j)  Tq,  the  tension  at  Q,  acting  along  the  tangent  to  the  string 
at^; 

(c)  the  reaction  with  the  surface. 

By  Lami's  theorem,  each  force  must  be  proportional  to  the  sine 
of  the  angle  between  the  remaining  two  forces. 


STRINGS 


75 


Let  A  be  the  point  of  the  surface  at  which  the  string  leaves  it. 
Let  the  normals  to  the  surface  be  drawn  at  A,  P,  Q,  and  let  the 
normal  at  P  make  an  angle  6  with  the  normal  at  A.  If  the  points 
A,  P,  Q  come  in  tliis  order,  as  in  fig.  40,  the  normal  at  Q  will  make 
with  the  normal  at  A  an  angle  slightly  greater  than  6,  —  say  6  +  dd, 
—  so  that  dd  is  the  small  angle  between  the  normals  at  P  and  Q. 


Fig.  40 

With  this  notation  the  angle  between  the  tensions  Tp  and  Tq  is 
TT  —  dd.  Let  the  angle  between  the  reaction  R  and  the  tension  Tp 
be  a,  then  the  angle  between  the  tension  T^  and  R  iqtt  —  a  +  dd. 
Thus  we  have 


R 


T. 


T 
—        Q 


sin(7r  —  dd)       sin(7r  —  a  +  dd)       sin  a 
Since  sin(7r  —  a  +  dd)  =  sin  (a  —  dd),  we  have 


_2_ 


^r 


sin  a      sin  (a  —  dd) 
and  by  a  known  theorem  in  algebra,  each  fraction  is  equal  to 


sin  a  —  sin  {a  —  dd) 

Now  Tg  —  Tp  is  the  increase  in  T  when  d  changes  from  d  to  d  +  dd, 
and  this,  in  the  notation  of  the  differential  calculus,  may  be  written 

dT 


dd 


dd. 


7G  STATICS   OF   SYSTEMS  OF  PARTICLES 

Also  the  denominator  sin  a  —  sin  (a  —  dO)  is  the  increase  in 

sin  a  when  a  changes  from  a  —  dO  to  a,  and  this  in  the  same  way 

is  equal  to 

c?(sina)  ^ 

— ^^- dU     or     cos  a  do. 

da 

Thus  the  original  fraction  is  equal  to 

dT 

dd  dT 

or     sec  a 


cos  a  dd  dd 

T  dT 

Thus  ■ — ^  =  sec  a  —~> 

sm  a  do 

or  r^  =  tannr^.  (12) 

When  in  the  limit  the  particle  P^  is  supposed  to  become  van- 
ishingly  small,  Tp  and  Tq  become  indistinguishable.  Let  us  denote 
either  by  the  single  letter  T,  so  that  T  is  now  simply  the  tension 
at  a  point  at  which  the  normal  makes  an  angle  6  with  that  at  A. 
If  the  strmg  is  just  on  the  point  of  slipping  in  the  direction  APQ, 
the  angle  between  the  reaction  R  and  the  normal  of  either  Q  or  P 
will  be  e,  the  angle  of  friction.    Thus  we  shall  have 

TT 

so  that  tan  a  =  cot  e,  and  equation  (12)  becomes 

r=cote^.  ■  (13) 

>•     54.  If  the  contact  between  the  surface  and  the  string  is  per- 

''  dT 

fectly  smooth,  e  =  0,  so  that  -t;^  =  0.    It  follows  that  T  is  a  con- 
•^  dd 

stant ;  i.e.  the  tension  is  the  same  at  all  points  of  the  string.    Thus 

/    the  tension  of  a  string  is  not  altered  by  its  passing  over  a  smooth 

surface,  —  the  result  already  given  in  §  36. 


DR.  GEORGE  F.  McEWEN 

STKINGS  77 

55.  In  the  more  general  use  in  which  the  contact  is  not  perfectly 
smooth,  let  /x  be  the  coefficient  of  friction,  so  that  fi  =  tan  e,  then 
equation  (13)  may  be  written 

dS      '^  ' 

dT 

and  integrating  this,  —  =  d  (fiO), 

d  (log  T)  =  d  i/xd), 
or  log  T  =  fi6  +  a  constant. 

Let  T^  be  the  tension  at  A,  then  we  find,  by  putting  6  =  0,  that 
the  constant  must  be  equal  to  log  T^,  so  that 

or  T  =  T  e^^  (14) 

If  the  string  leaves  the  surface  again  at  some  point  B  at  which 

the  normal  makes  an  angle  yjr  with  the  normal  at  A,  we  find  for 

the  tension  at  B 

T  =  Te^''', 

so  that  the  tension  is  multiphed  by  c'^'^  on  passing  over  the  surface 
from  A  to  B. 

If  the  string  (or  rope)  is  passed  round  and  round  a  post  or  bollard, 
the  tension  is  increased  in  the  ratio  e"**"  for  each  complete  turn. 
For  a  hemp  rope  on  oak  the  coefficient  of  friction,  according  to 
Morin,  is  fi  =  0.53.  Thus  2/A7r  =  3.34  and  c''^"=  28.1.  The  tension 
of  a  rope  wound  rovmd  an  oak  post  is  accordmgly  increased  about 
twenty-eight  fold  for  each  complete  turn. 

EXAMPLES 

^  1.  A  weight  is  suspended  by  a  rope  whicli,  after  being  wound  round  a  hori- 
zontal beam,  leaves  the  beam  horizontally,  its  end  being  controlled  by  a  work- 
man. If  the  rope  makes  li-  complete  revolutions  round  the  beam,  what  force 
must  be  exerted  by  the  man 

(a)  to  keep  the  weight  from  slipping  ? 

(6)  to  raise  the  weight  ? 

(Take  m  =  i-) 


78  STATICS  OF   SYSTEMS  OF  PARTICLES 

2.  A  weight  of  2^  pounds  stands  on  a  rough  table.  A  string  tied  to  the 
base  of  the  weight  hangs  over  the  edge  of  the  table  and  has  attached  to  it 
a  second  weight  which  hangs  freely.  If  the  coefficients  of  friction  between 
the  weight  and  the  table  and  the  string  and  the  table  are  h  and  ^  respec- 
tively, find  how  heavy  the  hanging  weight  must  be  to  start  the  other  weight 
into  motion. 

3.  A  weight  of  2500  pounds  is  to  be  raised  from  the  hold  of  a  ship.  A  rope 
attached  to  the  weight  makes  3}  turns  round  a  steam  windlass,  its  other  end 
being  held  by  a  seaman.  With  what  force  must  he  pull  his  end  of  the  rope  so 
as  to  raise  the  weight  when  the  windlass  is  in  motion  ?    {n  =  |.) 

4.  In  the  last  question,  find  what  pull  would  have  to  be  exerted  on  the  rope 
if  the  windlass  were  at  rest. 

5.  It  is  found  that  two  men  can  hold  a  weight  on  a  rope  by  taking  three 
turns  about  a  post,  and  that  one  of  them  can  do  it  alone  by  taking  one  half 
turn  extra.  If  each  can  pull  with  a  force  of  220  pounds  weight,  find  the  weight 
sustained. 

6.  In  a  tug  of  war  the  rope  is  observed  to  rub  against  a  post  at  the  critical 
moment,  in  such  a  way  that  the  two  parts  of  the  rope  make  an  angle  of  1°  with 
one  another.  If  the  coefficient  of  friction  between  the  rope  and  the  post  is  |, 
show  that  this  imposes  a  handicap  on  the  winning  side  equal  to  about  .0029 
times  its  aggregate  pull. 

Suspensio7i  Bridge 

56.  An  interesting  problem  is  afforded  by  the  kind  of  suspen- 
sion bridge  in  which  the  weight  of  the  bridge  (supposed  horizontal) 
is  taken  by  a  suspension  cable  by  means  of  vertical  chauis  con- 
necting the  bridge  with  the  cable. 
Let  us,  for  simplicity,  agree  to 
neglect  the  weights  of  the  chains 
and  cable,  and  suppose  the  weight 
of  the  bridge  to  be  distributed 
evenly  along  its  length. 

Let  0  be  the  lowest  point  of 
the  cable,  and  let  P  be  any  other 
point.  Let  o,  p  be  the  points  of  the  bridge  vertically  below  O,  P, 
and  let  o})  —  x.  Let  the  tension  at  P  be  T,  and  let  that  at  0  be  H. 
Let  the  direction  of  the  cable  at  P  make  an  angle  d  with  the 
horizontal. 


SUSPENSION  BKIDGE  79 

The  forces  acting  on  the  piece  OP  of  the  cable  consist  of 

(a)  the  tension  at  0,  of  amount  if  acting  horizontally; 

(b)  the  tension  at  P,  of  amount  T  acting  at  an  angle  6  with 
the  horizontal; 

(c)  the  tensions  of  the  vertical  chains,  all  acting  vertically. 

Eesolving  horizontally,  we  obtain 

II-Tcosd  =  0.  (15) 

Eesolving  vertically,  we  obtain 

T  sm  6*  -  ,S'  =  0, 

where  S  is  the  sum  of  the  tensions  of  all  the  chains  which  leave 
the  cable  between  0  and  P.  These  tensions  support  the  portion 
oj}  of  the  bridge,  and  if  the  weight  of  the  bridge  is  tv  per  unit 
length,  the  weight  of  ojy  will  be  ivx.    Thus  S  =  wx,  and  therefore 

Tsin6  =  ivx.  (16) 

This  and  equation  (15),  Tcos  6  =  IT,  (17) 

will  give  us  all  the  information  we  require. 

To  find  the  shape  which  the  cable  must  have  in  order  that  the 
bridge  may  hang  horizontally,  we  require  to  obtain  a  relation 
between  0  and  x.    We  accordingly  eliminate  T  from  equations  (16) 

and  (17),  and  obtain 

tan  6  —  —  X. 
H 

If  y  is  the  height  of  the  cable  above  the  bridge,  x,  y  may  be 
regarded  as  the  Cartesian  coordinates  of  a  pomt  P  on  the  cable, 

and  we  have  ■, 

tan^  =  ^. 
ax 

Thus  the  coordmates  x,  y  oi  P  are  related  by 

dy       w 

-^  =  —  .r, 
dx       U 

giving  on  integration  y  =  -  —  oi?  +  G, 

2  H 

where  C  is  a  constant  of  intesfration. 


80 


STATICS  OF  SYSTEMS  OF  PARTICLES 


Tliis  is  the  Cartesian  equation  of  the  cable.    It  is  easily  seen  to 

211 
represent  a  parabola  of  latus  rectum 


Thus  the  cable  must 


liang  in  the  form  of  a  parabola.  The  greater  the  horizontal  ten- 
sion, the  greater  the  latus  rectum  of  the  parabola,  and  therefore 
the  flatter  the  curve  of  the  cable.  A  perfectly  straight  cable  is  of 
course  an  impossibility  —  this  would  require  infinite  tension. 

57.  To  find  the  tension  at  any  point  of  the  cable,  we  square 
equations  (16)  and  (17),  and  add  corresponding  sides.    Thus 


giving  the  tension  at  a  point  distant  x  from  the  center, 
bridge  is  of  length  2  a,  the  tension  at  either  pier  must  be 


If  the 


^H-  +  %ohi^ 


^  V 


The   Catenary 

58.  In  the  problem  of  the  suspension  bridge,  we  neglected  the 
weight  of  the  cable.  A  second  problem  arises  when  the  cable  is 
svipposed  to  be  acted  on  by  no  external  forces  except  its  own 
weight.  The  problem  here  is  simply  that  of 
a  string  of  which  the  two  ends  are  fastened 
to  fixed  points,  and  which  hangs  freely  be- 
tween these  points. 

As  before,  let  0  be  the  lowest  point,  and 
let  P  be  any  other  point.  The  forces  acting 
on  the  portion  OP  of  the  string  are 

(«)  the  tension  at  0,  of  amount  H  acting 
horizontally ; 

(Ij)  the  tension  at  P,  of  amount  T  acting 
at  an  angle  d  with  the  horizontal ; 
(c)  the  weight  of  OP.    If  we  take  the  string  to  be  of  weight  w 
per  unit  length,  and  denote  the  distance  OP  by  s,  this  weight  is 
%os  acting  vertically. 


THE  CATENARY  81 

Eesolving  horizontally,  we  obtain 

JI-Tcose  =  0.  (18) 

Eesolving  vertically,  we  obtain 

T  sin  6  — us  =  0.  (19) 

To  find  the  shape  of  curve  in  which  the  string  will  hang,  we 
must  obtain  a  relation  between  6  and  s.    Eliminating  T,  we  obtain 

H  tan  6  =  ws, 

or,  if  we  replace  J£/iv  by  the  single  constant  c, 

s  =  c  tan  6.  (20) 

This  is  one  form  of  the  equation  of  the  curve,  s  and  d  being  taken  as 
coordinates.  The  equation  in  this  form  is  known  as  the  intrinsic  equation 
of  the  curve.  We  require,  however,  to  deduce  the  equation  in  its  Carte- 
sian form. 

59.  If  the  point  o  in  fig.  42  is  taken  as  origin,  the  axes  being 
horizontal  and  vertical,  we  have  at  once  the  rela- 

dx:dy:ds  =  cos  6  :  sin  6:1,  (2 1) 

for  dx  and  di/  are  the  horizontal  and  vertical  pro- 
jections of  the  small  element  ds  of  length  of  the 
string.  As  a  first  step,  let  us  use  relations  (21)  to 
change  the  variables  of  equation  (20)  from  s  and  0  to  s  and  y. 

We  have  ^  ^  ^  ,^^.2 

c^  =  s^  cot^  6  —  s'  cosec"  6  —  s-  =  s^  i  -—]  —  s^, 

,1     ,  ds  /— r, 

so  that  s  —  =  vs-  -(-  c". 

di/ 

Ihus  di/ =  —  > 

and  integrating  this,  we  obtain 


1/  =  Vi'-  +  c^  +  a  constant.  (22) 

We  can  determine  the  constant  of  integration  as  soon  as  we 
decide  where  the  origin  is  to  be  taken  —  so  far  we  have  not  fixed 


82  STATICS   OF   SYSTEMS   OF  PAKTICLES 

the  point  o.  Since  s  denotes  the  arc  of  the  curve  measured  from 
0,  we  have  s  =  0  at  the  pomt  0,  and  therefore  the  y  coordinate 
of  0  (putting  s  =  0  in  equation  (22))  is 

y  =  c  +  a  constant. 

Let  us  agree  that  Oo  is  to  be  made  equal  to  c,  so  that  y  =  c  at  0. 
Then  the  unknown  'constant  of  integration  must  be  zero.  Thus 
ecjuation  (2  2)  will  be         f  =  s^  +  c\  (2 3) 

The  last  step  is  to  transform  the  variables  from  y  and  stoij  and  x. 
The  relation  which  enables  us  to  do  this  is  obtained  by  eliminating 
0  from  relations  (21),  and  is 

{dsy  =  {dyf  +  {dxf.  (24) 

The  equation  already  obtained  is 

so  that  ds  =    /    ^     > 

V  7/^  —  c^ 

and  on  eliminating  ds  from  this  and  relation  (24),  we  obtain 

y^pl=^dyf  +  {dxf. 

y'  —  c 


From  this,  {dxf  =  {dyf 


f 


y'-c' 


-1 


SO  that  dx  =  ^     •  (25) 

V  y'^  —  c^ 

Integrating  this,  we  obtain 

-  =  cosh->  (26) 

c  c 

X  X 

where  cosh  -  =  ^  (e"  +  e  ''). 

c 

The  student  who  is  not  familiar  with  the  hyperbolic  cosine  (cosh)  fiuic- 
tion  will  easily  be  able  to  verify  equation  (26)  by  differentiating  it  back 
into  equation  (25). 


THE  CATENARY  83 

Equation  (26)  is  the  Cartesian  equation  of  the  curve  formed  by 
the  string;  tliis  curve  is  known  as  the  catenary. 

From  equation  (23)  we  obtain  the  value  of  s  in  the  form 


s 

=  y-c 

=  c^l  cosh^  — 

\ 

=  c^  sinh^  -  > 

c 

s 

=  sinh  -> 

c 

c 

imh- 
c 

=  l{^-e~S. 

so  that 
where 


60.  Expanding  the  exponentials   e",  e  ",  we  obtain  cosh  -   in 

the  form  i   /  \  2       1    /  \4 

.   X      ^       1    x\  ^       1     x\ 

So  long  as  x  is  small,  we  may  neglect  all  the  terms  of  this  series 
beyond  the  second.  Using  the  value  obtained  in  this  way,  we 
obtain  instead  of  equation  (26) 

x^ 
^  2c 

showing  that  so  long  as  x  is  small  the  curve  coincides  very  approxi- 
mately with  a  iKivahola  of  latus  rectum  2  c  or  2  IT  lie. 

This  parabola,  it  will  be  noticed,  is  one  which  would  he  formed  by  the 
cable  of  a  suspension  bridge  of  horizontal  tension  //,  w  being  the  weight 
per  unit  length  of  the  bridge  itself.  Indeed,  it  is  clear  that  when  the  cable 
is  almost  horizontal,  it  is  a  matter  of  indifference  whether  the  cable 
itself  possess  weight  w  per  unit  length  of  its  arc,  or  whether  a  weight  w 
per  unit  length  is  hung  from  it  so  as  to  lie  horizontally. 

We  can  also  obtain  a  simple 'approximation  to  the  sliape  of  the 
catenary  when  x  is  large,  i.e.  at  points  far  removed  from  the  lowest 

pomt.    When  x  is  verv  large,  so  that  -  is  verv  large,  the  value  of 
e"  becomes  very  large,  while  that  of  e  "  becomes  very  small.     Thus 


84 


STATICS   OF   SYSTEMS  OF  PARTICLES 


the  value  of  cosh  -  becomes  approximately  ^  c%  and  the  equation 
of  the  catenary  (equation  (26))  becomes 

X 


Thus   for    large  values   of    x   the    catenary   coincides   with   the 
exponential  curve. 


1 

V                                      1 

7c                                                / 

\ 

6c                                             / 

\ 

5C                                        / 

\\ 

4C                                    /     / 

^ 

3C                             1  / 

\    - 

"  J 

_^ 

^ 

o 

Fig.  44 


2e 


Fig.  44  shows  the  form  of  the  catenary.    The  thin  curves  are 

(a)  the  parabola,  to  which  the  catenary  approximates  for  small 
values  of  x\ 

(b)  the  exponential  curve,  to  which  the  catenary  approximates 
for  large  values  of  x. 


THE  CATENARY 


85 


61.  Sag  of  a  tightly  stretched  string.  A  string  or  wire  stretched 
so  as  to  be  nearly  liorizontal  all  along  its  length  —  as  for  instance 
a  telegraph  wire  —  may,  as  we  have  seen,  be  supposed  to  form  a 
parabola  to  within  good  approximation.  Thus  let  A,  B  be  two  poles 
at  equal  height  between 
which  a  wire  is  stretched ; 
let  C  be  the  middle  point 
of  AB,  and  let  D  be  the 
point  of  the  wire  verti- 
cally below  C.  Then,  Fig.  45 
from  symmetry,  D  will  be  the  lowest  point  of  the  wire  and  there- 
fore will  be  the  vertex  of  the  parabola.  Thus,  from  the  equation 
of  the  parabola,                             o  tt 


CB^ 


CD, 


since,  by  §  60,  its  latus  rectum  is  2  H/iv. 

Thus  if  h  =  AB,  the  distance  between  the  poles,  the  "  dip  "  CD  is 

given  by 


CD  = CB^ 

2H 

~8  if  ' 


(2") 


To  obtain  the  length  of  the  wire  we  have  to  introduce  a  higher 
order  of  small  quantities,  and  so  are  compelled  to  return  to  the 
equation  of  the  catenary. 


We  have 


^jcic'  —  e 

1  iC' 


=  X  -\- 


6c'^ 


) 

+  ■ 


The  quantity  we  require  is  s  —  x,  namely  DB  —  CB  in  fig.  45. 
When  the  string  is  tightly  stretched  c  is  very  great,  so  that  we 
may  neglect  the  terms  in  s  beyond  those  written  down  in  the 

above  equation,  and  obtain 

1 


s  —  X  =■ 


6  c 

1  VJ^X^ 


approximately, 


DR.  GEORGE  F.  McEWEN 

86      STATICS  or  SYSTEMS  OF  PARTICLES 

Putting  X  =  l^  h,  we  find  as  tlie  total  increase  in  a  span  of 
length  h,  caused  by  sagging, 

1   ^t'%« 

EXAMPLES 

1.  The  entire  load  of  a  suspension  bridge  is  320  tons,  the  span  is  040  feet, 
and  the  height  is  50  feet.  Find  tlie  tension  at  the  points  of  support,  and  also 
the  tension  at  the  lowest  point. 

2.  The  weight  of  a  freely  suspended  cable  is  320  tons ;  the  distance  between 
the  two  points  of  support,  which  are  in  the  same  horizontal  line,  is  640  feet, 
and  the  height  of  these  points  above  the  lowest  point  of  the  cable  is  50  feet. 
Find  the  tension  at  the  points  of  support,  and  also  the  tension  at  the  lowest  point. 

3.  The  wire  for  a  telegraph  line  cannot  sustain  a  weight  of  more  than  a  mile 
of  its  own  length  without  breaking.  If  the  wire  is  stretched  on  poles  at  equal 
intervals  of  88  yards,  what  is  the  least  sag  permissible  ? 

4.  In  the  last  question  how  much  wire  is  required  for  a  mile  of  the  line  ? 

5.  A  telegraph  line  has  to  be  built  of  a  certain  kind  of  wire,  stretched  over 
evenly  spaced  posts.  Show  that  if  the  number  of  posts  is  very  large,  the  line 
will  be  built  most  economically  as  regards  the  cost  of  wire  and  posts,  if  the 
cost  of  the  posts  is  three  times  that  of  the  additional  length  of  wire  required  by 


GENERAL  EXAMPLES 

1.  A  block  of  stone  weighing  J  ton  is  raised  by  means  of  a  rope 
which  passes  over  a  pulley  vertically  above  it  and  is  w^ound  upon  a  wind- 
lass one  foot  in  diameter.  The  windlass  is  worked  by  two  men  who  turn 
cranks  of  length  2  feet.  What  force  must  each  man  exert  perpendicular 
to  the  cranks  ? 

2.  A  man  sitting  in  one  scale  of  a  balance  presses  with  a  force  of  60 
pounds  against  the  beam  in  a  vertical  direction  and  at  a  point  halfway 
between  the  fulcrum  and  the  end  of  the  beam  from  which  his  scale  is  sup- 
ported. If  the  beam  is  5  feet  long,  find  the  additional  weight  which  must 
be  put  in  the  other  scale  to  maintain  equilibrium. 

3.  The  scales  of  a  false  balance  hang  at  unequal  distances  a,  b  from 
the  fulcrum,  but  balance  w^hen  empty.  A  weight  appears  to  have  weights 
P,  Q  respectively  when  w^eighed  in  the  two  scales.  Find  its  true  weight 
and  prove  that 

b         [Q 


EXAMPLES  87 

4.  A  weightless  string  24  inches  in  length  is  fastened  to  two  points 
which  are  in  the  same  horizontal  line,  and  at  a  distance  of  16  inches  apart. 
Weights  are  fixed  to  two  points  at  distances  of  9  and  7  inches  from  the 
ends  of  the  string  and  hang  in  such  a  way  that  the  portion  of  the  string 
between  them  is  horizontal.    Determine  the  ratio  of  the  weights. 

5.  A  light  wire  has  a  weight  suspended  from  its  middle  jioint,  and  is 
itself  supjiorted  by  a  string  fastened  to  its  two  ends  and  passing  over  a 
smooth  peg.  Show  that  the  wire  can  rest  only  in  a  horizontal  or  vertical 
position. 

6.  Three  smooth  pegs  A ,  B,  C  stuck  in  a  wall  are  the  vertices  of  an 
equilateral  triangle,  A  being  the  highest  and  the  side  BC  horizontal  ;  a 
light  string  passes  once  around  the  pegs  and  its  ends  are  fastened  to  a 
weight  W  which  hangs  in  equilibrium  below  BC.  Find  the  pressure  on 
each  peg. 

7.  Two  rings  of  weights  P  and  Q  respectively  slide  on  a  weightless 
string  whose  ends  are  fastened  to  the  extremities  of  a  straight  rod  inclined 
at  an  angle  6  to  the  horizontal.  On  this  rod  slides  a  light  ring  through 
which  the  string  passes,  so  that  the  heavy  rings  are  on  different  sides  of 
the  light  ring.  All  contacts  are  smooth  and,  in  equilibrium,  (p  is  the  angle 
between  the  rod  and  those  parts  of  the  string  which  are  close  to  the  light 
ring.    Prove  that  Un9  _P-Q 

tan  4>~  P  +  Q' 

8.  Two  small  heavy  rings  slide  on  a  smooth  wire,  in  the  shape  of  a 
parabola  with  axis  horizontal  ;  they  are  connected  by  a  light  string  wliich 
passes  over  a  smooth  peg  at  the  focus.  Show  that  their  depths  below  the 
axis  are  proportional  to  their  weights  when  they  are  in  equilibrium. 

9.  Two  equally  heavy  rings  slide  on  a  wire  in  the  shape  of  an  ellipse 
whose  major  axis  is  vertical,  and  are  connected  by  a  string  which  i^asses 
over  a  smooth  peg  at  the  upper  focus.  Show  that  thei-e  are  an  infinite 
number  of  positions  of  equilibrium. 

-  /  10.  A  BCD  is  a  quadrilateral;  forces  act  along  the  sides  AB.  BC.  CD, 
DA  measured  by  a,  /3,  y,  5  times  those  sides  respectively.  Show  that  if 
these  forces  keep  any  system  of  particles  in  equilibrium,  then 

^^  yll.  A  light  rod  rests  wholly  within  a  smooth  hemispherical  bowl  of 
radius  r,  and  a  weight  W  is  clamped  on  to  the  rod  at  a  point  whose  dis- 
tances from  the  ends  are  o  and  b.  Show  that  d,  the  inclination  of  the  rod 
to  the  horizon  in  the  position  of  equilibrium,  is  given  by  the  equation 

2  Vr2  —  ab  sin  5  =  a  -  b. 


88  STATICS   OF   SYSTEMS   OF  PAFtTlCLES 

12.  A  weightless  rod,  to  which  are  fixed  two  rough  beads  of  masses  m 
and  m' ,  lies  on  an  inclined  plane  and  is  free  to  turn  aliout  an  axis  through 
the  rod  periiendicular  to  the  plane.  Tf  it  be  in  a  horizontal  position,  show 
that  it  will  not  begin  to  slip  round  unless 

nm  ^^  ui'l)  , 

/a  < —  tan  a, 

iiKi  +  iii'b 

where  a  is  the  angle  of  the  plane,  and  a  and  h  the  distances  of  m  and  m' 
re.spectively  from  the  axis. 

13.  A  bead  of  weight  IF,  run  on  a  smooth  weightless  string,  rests  on  an 
inclined  plane  of  angle  a,  the  coefficient  of  friction  between  the  bead  and 
jilane  being  /x.  The  ends  of  the  string  are  tied  to  two  points  A,  B  iu  the 
])lane  at  the  same  height.  Show  how  to  find  the  positions  of  limiting- 
equilibrium  for  the  bead,  and  show  that  in  such  a  jsosition  P,  the  tension 

of  the  strincr  is  , 

°  i  W  sec  i  .4  PB  ■  (tan-^  a  -  ij?) "-. 

14.  A  uniform  string  is  placed  on  a  rough  sphere  so  as  to  lie  on  a  hori- 
zontal small  circle  in  altitude  a.  Prove  that,  if  the  string  be  on  the  point 
of  slif)ping  along  the  meridians,  the  tension  is  constant  and  equal  to  W  cot 
(<T  +  e),  where  TF  is  the  weight  of  a  length  of  the  string  ecpial  to  the  radius 
of  the  circle,  and  e  is  the  angle  of  friction. 

15.  A  weightless  string  is  suspended  from  two  fixed  pf)ints  and  at  given 
points  on  the  string  equal  weights  are  attached.  Prove  that  the  tangents 
of  the  inclinations  to  the  horizon  of  different  i^ortions  of  the  string  form 
an  arithmetical  progression. 

16.  A  smooth  semicircular  tube  is  just  filled  with  2  n  equal  smooth 
beads,  each  of  weight  W,  that  just  fit  the  tube,  and  stands  in  a  vertical 
plane  with  the  two  ends  at  equal  height.  If  R,„  is  the  pressure  between 
the  ?Hth  and  (m  +  l)th  beads  from  the  top,  show  that 

Ji,,,  =  ]V  sni  —  cosec 

2  «  2  n 

17.  In  the  last  question,  let  the  beads  be  indefinitely  diminished  in 
size.  Prove  that  the  pressure  between  any  two  beads  is  projwrtional  to 
the  depth  below  the  top  of  the  tube. 

18.  A  heavy  string  hangs  over  two  smooth  pegs,  at  the  same  level  and 
distance  a  apart,  the  two  ends  of  the  strings  hanging  freely  and  the  central 
part  hanging  in  a  catenary.  Show  that  for  equilibrium  to  be  possible,  the 
total  length  of  the  string  must  not  be  less  than  ae. 


EXAMPLES  80 

19.  A  string  of  weight  W  is  suspended  from  two  points  at  the  same 
level,  and  a  weight  W  is  attached  to  its  lowest  point.  If  a,  /3  are  now  the 
inclinations  to  the  vertical  of  the  tangents  at  the  highest  and  lowest  points, 
prove  that  ^^^^  ^  ^      ^ 

tan/i~  IF'" 

20.  A  heavy  string  of  length  I  is  supported  from  two  points,  and  at 
these  points  the  string  makes  angles  a,  ji  with  the  vertical.  Show  that 
the  height  of  one  point  above  the  other  is 

I  cos  },  («  +  ^3)  sec  ^  («  —  /3). 

21.  Prove  that  the  direction  of  the  least  force  required  to  draw  a 
carriage  is  inclined  at  an  angle  0  to  the  ground,  where  asin^  =  isine, 
a  and  b  being  the  radii  of  wheels  and  axles  respectively,  and  e  being  the 
angle  of  friction. 


CHAPTER  V 
STATICS  OF  RIGID  BODIES 

ElGIDITY 

62.  If  we  press  a  lump  of  wet  clay  or  of  soft  putty  with  the 
finger,  we  fiud  that  a  dent  is  left  in  the  clay  or  putty ;  the  force 
applied  to  the  substance  by  the  finger  has  caused  it  to  change  its 
shape.  If  we  press  a  mass  of  jelly  with  the  linger,  we  do  not  find 
any  dent  left  in  the  jelly,  but  we  notice  that  so  long  as  the  force 
is  applied  the  shape  of  the  jelly  is  altered,  although  it  returns  to 
its  original  shape  as  soon  as  the  pressure  is  removed. 

On  the  other  hand,  if  we  press  a  lead  bidlet  or  an  ivory  biQiard 
ball  with  the  finger,  we  do  not  notice  any  change  of  shape  either 
while  the  pressure  is  applied  or  after.  In  ordinary  language  we 
say  that  the  lead  and  ivory  are  harder  ilmn  the  clay  and  putty; 
in  scientific  language  we  say  that  they  are  more  rigid. 

63.  A  perfectly  rigid  body  would  be  one  which  showed  no 
change  of  shape  under  any  force,  no  matter  how  great  this  force 
might  be.  A  bullet  and  a  billiard  ball  are  not  perfectly  rigid.  A 
billiard  ball  is  pressed  out  of  shape  during  the  interval  of  collision 
with  a  second  billiard  ball,  but  regains  its  shape  immediately, 
while  a  lead  bullet  is  pressed  permanently  out  of  shape  by  strikmg 
a  target.  A  perfectly  rigid  body  does  not  exist  m  nature,  although 
such  bodies  as  a  billiard  ball  or  a  liullet  may  be  regarded  as  per- 
fectly rigid,  so  long  as  the  forces  wliich  act  upon  them  are  not 
too  great. 

We  can  give  a  mathematical  definition  of  a  perfectly  rigid  body 
as  follows : 

A  hody  is  'perfectly  rigid  ivhen  the  distance  between  any  pair 
of  particles  in  it  remains  unaltered,  no  matter  what  forces  act  on 
the  hody. 

90 


RIGIDITY  91 

64.  A  rigid  body  can  move  about  in  space  without  changing 
the  direction  of  any  line  in  it  —  such  a  motion  is  called  a  motion 
of  translation.  It  can  also  turn  about  any  pomt  P  without  the 
position  of  P  altering  —  such  a  motion  is  called  a  motion  of  rota- 
tion about  P.  It  can  again  have  a  motion  compounded  of  a  motion 
of  translation  and  one  of 'rotation,  and  this  we  shall  show  to  be 
the  most  general  motion  wliich  it  can  possibly  have. 

65.  First  we  must  notice  that  a  rigid  body  is  fixed  when  any 
three  points  in  it  are  fixed,  provided  that  these  three  points  do  not 
lie  in  a  straight  Hne.  For  let  A,  B,  C  ho.  the  three  points.  If  we 
fix  A  and  B,  any  motion  wliich  can  take  place  must,  since  the 
body  is  by  hypothesis  perfectly  rigid,  be  one  in  which  the  dis- 
tances of  any  other  point  P  from  A  and  B  remam  unaltered. 
Thus  P  must  describe  a  circle  with  AB  as  axis,  and  the  motion 
of  the  body  must  be  one  of  rotation  about  the  line  AB.  Thus 
if  A,  B,  C  are  not  in  a  straight  Ime,  C  must  describe  a  circle 
about  AB.  But  if  C  also  is  fixed,  this  cannot  happen;  in  other 
words,  no  motion  can  take  place,  so  that  the  body  is  fixed  in 
position. 

Thus  the  position  of  a  rigid  body  is  determined  when  the  posi- 
tions of  three  non-collinear  points  in  it  are  determined. 

66.  We  can  now  prove  the  theorem : 

The  most  general  motion  of  a  rigid  hody  is  compounded  of  a 
motion  of  translation  and  one  of  rotation. 

In  fig.  46  let  the  figure  on  the  left  represent  the  body  in  its 
original  position,  and  let  the  heavy  curve  on  the  right  represent 
the  body  after  it  has  been  moved  m  any  way.  Let  P  be  the 
position  of  any  particle  of  the  body  in  its  original  position,  and 
let  Q  be  the  position  of  the  same  particle  after  the  motion  has 
taken  place. 

Imagine  that  the  body  is  first  moved  from  its  original  position 
in  such  a  way  that  the  point  P  moves  to  Q,  while  all  lines  in  the 
body  remain  parallel  to  their  original  positions.  This  motion  is 
one  of  pure  translation.  After  this  motion  has  taken  place,  we 
can  turn  the  body  about  the  point  Q  in  such  a  way  as  to  turn  it 


92 


STATICS   OF  RIGID   150DIES 


into  its  final  position.  For  let  us  take  any  two  other  particles  R,  S 
in  the  body  (not  in  the  same  straight  line  with  Q),  and  let  R',  S'  be 
their  final  positions.    Since  the  body  is  supposed  to  be  perfectly 


.-,•''' \ 


Fig.  46 

rigid,  it  follows  that  all  distances  between  particles  of  the  body 
remain  unaltered.  Hence  the  distances  QR,  RS,  SQ  are  respectively 
equal  to  QR',  R'S',  S'Q.  Thus  the  triangles  QRS,  QR'S',  being 
equal  in  all  respects,  can  be  superposed,  and  the  motion  of  super- 
position of  these  triangles  is  the  motion  required  and  is  a  motion 
of  pure  rotation  about  Q,  since  Q  does  not  move. 

Since  the  position  of  a  rigid  body  is  fixed  when  any  three  points 
in  it  are  fixed,  it  follows  that  the  rigid  body  can  only  have  one 
position  in  which  the  three  points  Q,  R,  S  have  given  positions.  But 
after  the  motion  we  have  described,  the  three  points  Q,  R,  S  are 
placed  in  their  finaj.  positions.  Hence  the  whole  body  must  be  in 
its  final  position,  and  this  proves  the  theorem.  ^^ 

67.  Axis  of  rotation. 
In  a  motion  of  rotation, 
let  P  1)6  the  point  which 
remains  fixed.  Take  any 
plane  A  through  P,  and 
let  B  be  the  position  of 
the  plane  A  after  the 
rotation  has  occurred. 
These   two   planes    both   pass    through    P,   and    must    therefore 


Fig.  47 


CONDITIONS   OF   EQUILIBRIUM  93 

intersect  in  some  line  PQ  passing  through  P.  This  line  is  called 
the  axis  of  rotation.  The  rotation  can  be  imagined  as  a  turning 
about  an  imaginary  pivot  rvmning  along  the  axis  of  rotation. 

Conditions  of  Equilibkium  for  a  Eigid  Body 

68.  AVe  have  seen  that  a  rigid  body  is  fixed  when  three  non- 
collinear  points  in  it  are  fixed.  It  follows  that,  whatever  forces 
act  on  a  rigid  body,  we  can  always  hold  it  at  rest  by  applying 
three  suitably  chosen  forces  at  three  points  which  are  not  in  a 
straight  line. 

AVe  can,  however,  select  these  forces  in  a  special  way. 

Let  us  select  any  three  points  A,  B,  C,  subject  only  to  tlie  con- 
dition that  they  are  not  in  a  straight  line.  By  a  suitably  chosen 
force  acting  on  the  particle  at  A  we  shall  always  be  able  to  keep 
the  point  A  at  rest. 

"VVlien  A  is  fixed  B  may  or  may  not  tend  to  move.  If  B  tends 
to  move,  the  direction  of  motion  of  B  must  be  perpendicular  to 
BA,  since  A  cannot  move.  Hence  after  A  is  fixed  it  must  be 
possible  to  fix  B  by  applying  at  7>  a  force  perpendicular  to  BA. 

AVhen  A  and  B  are  both  fixed  the  only  motion  possible  for  the 
third  point  C  is  one  perpendicular  to  both  AC  and  BC,  i.e.  perpen- 
dicular to  the  plane  ABC.  Thus  C  can  be  held  at  rest  by  a  force 
perpendicular  to  the  plane  ABC,  and  the  whole  body  is  now  held 
at  rest.  Thus  it  has  been  proved  that  a  rigid  hocbj  can  he  held  at 
rest,  in  opposition  to  the  action  of  any  system  of  forces,  by  the 
application  of  the  following  forces  at  three  arbitrarily  chosen 
points  A,  B,  C  not  in  a  straight  line : 

(a)  a  force  at  A,  direction  unknown ; 

(&)   a  force  at  B,  direction  per})endicular  to  the  line  AB ; 

(c)  a  force  at  C,  direction  perpendicidar  to  the  plane  ABC. 

The  condition  that  the  original  system  of  forces  shoidil  Imld 
the  body  in  equilibrium  is  of  course  that  no  additional  forces  are 
required  to  fix  the  body,  and  hence  that  the  three  forces  introduced 
at  the  points  A,  B,  C  should  each  vanish. 


94  STATICS  OF  RIGID  BODIES 

Tkansmissibility  of  Force 

69.  Consider  a  rigid  body  acted  on  by  two  forces  JFj,  11],  at 
two  points  A,  B,  these  forces  being  equal  in  magnitude  but  acting 
in  the  opposite  directions  AB,  BA. 

Either  the  rigid  body  will  be  in  equilibrium  under  the  action  of 
these  two  forces,  or  else  it  can  be  held  at  rest  by  three  forces  F^, 
P^,  Pq  at  the  points  A^  B,  and  any  third  point  C  not  in  the  line 
AB,  these  forces  being  in  the  directions  already  mentioned,  namely 
Pf,  being  perpendicular  to  ABC,  and  Pj^  perpendicular  to  AB. 


Fig.  48 

Let  these  forces,  if  necessary,  be  put  in  so  that  the  body  is  in 
equilibrium  under  the  action  of  tlie  forces  TFj,  TJ^,  P,,  Pj^,  P^.. 
The  body  bemg  in  equdibrium,  the  sum  of  the  moments  of  these 
forces  about  any  line,  or  of  their  components  in  any  direction,  must 
vanish,  by  §  50. 

Let  us  consider  what  is  the  sum  of  the  moments  about  the  line 
AB.  The  forces  W^,  Wj^,  P^,  P^  all  meet  this  line,  so  that  the 
moment  of  each  of  these  forces  vanishes.  Thus  the  sum  of  the 
moments  about  the  line  AB  consists  of  the  moment  of  the  single 
force  Pf.,  and  for  the  sum  of  these  moments  to  vanish,  the  moment 
of  Pf.  must  vanish.  Now  i^  is  perpendicular  to  the  line  AB,  and 
does  not  intersect  it,  so  that  the  moment  of  P^  can  vanish  only  if 
the  force  Pf.  is  itself  equal  to  zero.  This  means  that  no  force  is 
required  to  keep  the  body  from  turning  about  AB  as  axis  of 
rotation. 


TEANSMISSIBILITY  OF  FOliCE  95 

The  body  is  now  held  at  rest  by  the  two  forces  F^^,  Fj^,  and  is 
therefore  in  equihbrium  under  the  forces  P,,  Pj,,  IV^,  Wj,.  Tak- 
ing moments  about  a  line  through  A  perpendicular  to  AB,  we  find 
that  the  moments  of  i^,  W\,,  Tl^  Vanish,  so  that  in  order  that  the 
sum  of  the  moments  of  the  four  forces  taken  about  this  Ime  may 
vanish,  we  must  have  the  moment  of  i^  equal  to  zero,  and  there- 
fore Pj,^  itself  equal  to  zero.  Thus  the  only  force  required  to  keep 
the  body  at  rest  is  the  force  P^  at  A. 

A  condition  for  eciuilibrium  is  now  that  the  sum  of  the  com- 
ponents of  Wj^,  Wj,,  and  P^  shall  vanish  in  any  direction.  The 
components  of  W^  and  Wj,  are,  however,  equal  and  opposite,  so 
that  the  component  of  Pj  must  vanish  in  every  direction.  That 
is  to  say,  we  must  have  P,  =  0. 

It  has  now  been  proved  that  the  rigid  Ijody  is  in  equilibrium 
under  the  action  of  the  two  forces  TFj,  1]],. 

70.  Tliis  establishes  at  once  a  prmciple  known  as  the  transmissi- 
hility  of  force. 

The  effect  of  a  force  acting  on  a  rigid  body  dei^ends  on  its  mag- 
nitude and  on  the  line  along  which  it  acts,  hut  not  on  the  'particular 
•particle  in  this  line  to  which  it  is  apijlied. 

For,  let  the  same  force  be  applied  at  any  ^  *~ 

two  points  Q,  R  of  its  line  of  action.     An 

equal  and  opposite  force  at  R  can  neutralize  either  of  the  two 
forces,  which  are  therefore  equivalent; 

Composition  of  Foeces  acting  in  a  Plane 

71.  Suppose  that  we  have  two  forces  P,  Q  actmg  at  two  points 
A,  B  of  a  rigid  body,  it  bemg  supposed  that  the  two  lines  of  action 
of  these  forces  lie  in  one  plane.  Then  the  two  lines  of  action, 
produced  if  necessary  beyond  the  points  A,  B,  will  meet  in  some 
point  C. 

By  the  principle  of  the  transmissibility  of  force  it  is  imma- 
terial whether  the  force  P  acts  at  A  or  at  C;  let  us  suppose  it  to 
act  at  C.    In  the  same  way  let  us  suppose  the  force  Q  to  act  at  C 


96 


STATICS   OF   RIGID   BODIES 


instead  of  at  B.  Then  we  have  the  hody  acted  on  by  two  forces 
P,  Q  which  act  on  the  same  particle  C.  These  may  be  compounded, 
according  to  the  rules  explained  m  Chapter  III,  into  a  single  force 

li  acting  at  C.  Thus  we  can 
compound  forces  of  which 
the  lines  of  action  intersect, 
even  though  they  do  not  act 
on  the  same  particle. 

Having  compounded  two 
forces  into  a  single  force,  we 
may  compound  this  resultant 
with  any  third  force  which 
lies  in  the  same  plane  as  the 
two  original  forces,  and  in 
this  way  obtain  a  resultant 
Fig.  50  oi  three  forces,  and  so  on. 


Thus  any  number  of  forces  which  all  lie  in  one  plane  may  be 
compounded  int(3  a  single  force.  This  force  is  called  the  resultant  of 
the  original  force. 

72.  An  exception  arises  when  we  attempt 
to  compound  two  parallel  forces,  for  their 
lines  of  action  do  not  meet.  This  difficulty, 
however,  is  easily  surmounted.  Let  P,  Q 
be  the  two  forces  to  be  compounded,  and 
let  AB  be  any  line  cutting  their  lines  of 
action  in  A,  B.  Let  us  add  to  the  system 
F,  Q  two  forces  : 

(a)  a  force  II  acting  along  BA ; 
(h)  a  force  Jt  acting  along  AB. 


Fig.  51 


These  two  forces  being  equal  and  opposite 
can  be  introduced  without  producing  any 
effect.    On  compounding  the  first  with  P 

we  obtain  a  resultant  P'  acting  at  A,  and  on  compounding  the 
second  with  Q  we  obtain  a  resultant  Q'  acting  at  B.     Thus  the 


COMPOSITION  OF  FORCES  97 

original  forces  P,  Q  have  been  replaced  by  the  new  forces  P' ,  Q'. 
The  lines  of  action  of  these  forces,  however,  will  not  in  general  be 
parallel,  so  that  they  may  be  compounded  into  a  single  resultant 
force  acting  through  their  point  of  intersection. 

73.  Let  us  suppose  that  the  forces  originally  to  be  compounded 
were  R^,  R^,--,  and  that  these  have  been  compounded  into  a 
single  resultant  R.  Let  us  take  axes  x,  y  in  the  plane  in  which 
these  forces  act,  and  let  the  components  of  R^  along  these  axes  be 
X^,  1\,  those  of  i?2  being  X^,  Y„,  and  so  on.  Finally  let  the  com- 
ponents of  R  be  X,  Y. 

The  system  of  forces  which  consists  of  the  original  forces  i?j, 
R^,  ■  •  ■,  together  with  the  resultant  R  reversed,  constitutes  a  system 
in  equilibrium.    Eesolving  parallel  to  the  axes,  we  obtain 

a;  + a;  + a;  + x=  o, 

Y,  +  Y,+  Y,  + r  =  0. 

Thus  the  components  of  R  are  given  by  the  equations 
Y=  Y,+  i;+  Y,  +  .... 

The  magnitude  of  R  can  be  found  from  the  equation 

R^-  =  A'^  +  ^^ 

whde  the  angle  6  which  the  line  of  action  of  R  makes  with  the 
axis  of  X  can  be  found  from  the  equation 

Y 

tan  6  =  —' 

X 

To  obtain  the  position  of  the  line  of  action  of  R  we  use  the  fact 
that  the  sum  of  the  moments  of 

i?„  R„,  i?3,  •  •  •,  and  —  R 

taken  about  any  point  in  the  })lane  nnist  vanish.  This  gives  us 
the  moment  of  R  about  any  point,  and  hence,  since  we  know  the 
magnitude  and  direction  of  R,  we  can  iind  the  position  of  its  line 
of  action. 


98 


STATICS  OF  RIGID  BODIES 


ILLUSTRATIVE  EXAMPLE 

Forces  P,  Q,  -R  act  on  a  rigid  body,  all  the  forces  being  in  one  plane,_and  their 
lines  of  action  forming  a  right-angled  isosceles  triangle  of  sides  a,  a,  y/2a.  Find 
their  resultant. 

Let  ABC  be  the  triangle,  the  forces  P,  Q,  R  acting  along  BC,  CA,  AB 
respectively.  Take  C  for  origin  and  CB,  CA  for  axes  of  x,  y.  Let  the  resultant 
have  components  X,  Y.    Then,  resolving  along  Cx,  we  obtain 

R 


X  =  -  P  + 

and  similarly  resolving  along  Cy, 

V2 


V2 


Thus  the  resultant  is  of  magnitude  R  given  by 
R2  =  X2  +  F2  =  /-  P  +  ^y+  (l 


P\2 

V2/ 


=  p-^  +  Q^  +  P2  _  V2  P(P  +  Q). 
The  angle  0  which  it  makes  with  Cx  is  given  by    C 
Y  R-V2Q 


tan  6  = 


Fig.  52 


X 


R-V2P 


To  find  the  line  of  action,  we  express  that  the  moment  of  R  about  C  must  be 
equal  to  the  .sum  of  the  moments  of  P,  Q,  and  R.  If  p  is  the  perpendicular 
from  C  on  to  the  line  of  action  of  the  re.sultant  R,  this  gives 

Rp  =  R-SL, 

V2 

R    a 

so  that  p  = » 

R  V2 

and  this  determines  the  line  of  action  of  R. 


V 


EXAMPLES 

(All  forces  are  suiiihisimI  to  be  acting  on  rigid  bodies) 

1.  A  BCD  is  a  sijuare.  and  along  the  .sides  AB,  BC,  CD  there  act  forces  of 
1,  2,  3  pounds  respectively.  Determine  the  magnitude  and  line  of  action  of  the 
resultant. 

/''  2.  ABCD  is  a  square  and  forces  P,  Q,  R,  S  act  along  the  sides  ^7?.  BC,  CD, 
DA  re.spectively.  What  is  the  condition  that  their  resultant  shall  pass  through 
the  center  of  the  square  ? 

In  question  2,  what  is  the  condition 
)  that  the  resultant  shall  pass  through  A  ? 

(6)  that  it  .shall  pass  through  B? 

(c)  that  the  four  forces  shall  be  in  equilibrium  ? 


M 


PARALLEL  FORCES  99 

y^.  Forces  P,  Q,  R  act  along  the  sides  of  a  triangle  ABC,  and  their 
resultant  passes  through  the  centers  of  the  inscribed  and  circumscribed 
circles.     Prove  that 

P  ^ Q_ ^  R 

cos  B  —  cos  C      cos  C  —  cos  A      cos  A  —  cos  B 

5.  If  four  forces  acting  along  the  sides  of  a  quadrilateral  are  in  equi- 
librium, prove  that  the  quadrilateral  must  be  plane. 

6.  ABCD  is  a  plane  quadrilateral  and  forces  represented  by  AB,  CB,  CD, 
AD  act  along  the.se  sides,  respectively,  of  the  quadrilateral.  Show  that  if  there 
is  equilibrium,  the  quadrilateral  must  be  a  parallelogram. 

"-.,7.  If  a  quadrilateral  can  be  inscribed  in  a  circle,  prove  that  forces  acting 
along  the  four  sides  and  proportional  to  the  opposite  sides  will  keep  it  in  equi- 
librium. Show  also  that  the  converse  is  true,  namely  that  for  equilibrium,  the 
forces  must  be  proportional  to  the  opposite  sides. 

'  8.  A  quadrilateral  is  inscribed  in  a  circle,  and  four  forces  act  along  the 
sides,  and  are  inversely  proportional  to  the  lengths  of  the.se  sides.  Show  that 
the  resultant  has  for  line  of  action  the  line  through  the  intersections  of  pairs 
of  opposite  sides. 

9.  Forces  act  along  the  four  sides  of  a  quadrilateral,  equal  respectively  to 
a,  b,  c,  and  d  times  the  lengths  of  those  sides.    Show  that  if  there  is  equilibrium, 

ac  =  bd, 

and  that  the  further  conditions  necessary  to  insure  equilibrium  are  that  the 

ratios  a  :  b  and  b  :  c  shall  be  the  ratios  in  which  the  diagonals  are  divided  at 

their  points  of  intersection. 

0.   In  the  last  question  show  that  the  pei-pendicular  distances  to  the  first 

e,  from  the  two  points  of  the  quadrilateral  wdiich  are  not  on  that  side,  are  in 

he  ratio 

a{c  -b)  :  d{b-  a). 


Parallel  Forces 

74.  Let  us  use  the  method  just  explaiued,  to  deteriuine  the 
resultant  of  two  parallel  forces  F,  Q. 

Take  any  point  0  on  the  line  of  action  of  P  as  origin,  and  take 
tliis  Hue  of  action  of  P  for  axis  Oy,  as  in  fig.  53.  Let  the  resultant 
be  E,  components  X,  Y.    Then  resolving  we  obtain 

X=0, 
Y=F+Q, 

so  that  the  resultant  force  is  of  magnitude  F  +  Q  and  acts  parallel 


100 


DR  GEORGE  F.  McEWEN 

STATICS   OF  RIGID   BODIES 


y 


P+Qn 


-4l...., 


'^Q 


to  Oi/.    Let  us  suppose  that  its  distance  from  Oy  is  b,  that  of  Q 
being  denoted  by  a.    Then  taking  moments  about  0  we  have 

, ,        h  a  a  —  h 

so  that  —  = = > 

Q      P+Q         P 

showing  that  the  line  of  action 
divides  the  distance  between  P  and 
Q  in  the  ratio  Q  :  P. 

Thus  we  have  shown  that  the 
resultant  of  tiuo  imrallel  forces  P, 
Q  is  a  force  of  magnitude  P  +  Q, 
parallel  to  these  forces,  of  tvhich 
the  line  of  action  divides  the  dis- 
tance between  the  lines  of  action  of 
P  and  Q,  in  the  ratio  Q :  P. 

75.  Alternative  treatment  of  parallel  forces.    V\e  can  prove  directly  from 
§  68   that   parallel    forces    P,    Q    will    be    in    equilibrium    with    a   force 

—  (P+  Q)  parallel  to  them  and  acting  along  a  line  which  divides  the  dis- 
tance between  them  in  the  ratio  Q  :  P. 

Take  two  points  ^ ,  ^  on  the  line  of  action  of  P,  Q,  and  a  third  point 
C  not  on  the  line  AB.    Then  the  body,  acted  on  by  the  forces  P,  Q,  and 

—  (P  +  Q),  can  be  kept  in  equilibrium  by  the  further  application  of 
(a)  a  force  Rf^at  C,  perpendicular  to  ABC  ; 

(/;)  a  force  R^  at  B,  perpendicular  to  AB  ; 

(c)  a  force  R ^  at  A. 

Thus  the  system  of  forces 

P,  Q,  -(P  +  Q),  Re,  Rb,  Ra 
will  be  in  equilibrium. 

Taking  moments  about  the  line  AB,  we  find 
that  Rq=0.  Taking  moments  about  A  ,  we  find 
that  Rjj=0,  or  else  that  it  acts  along  BA,  in 


Fig.  0.3 


which  case  it  can  be  absorbed  into  R^.  Resolv- 
ing perpendicular  to  the  plane  of  the  forces  P,  Q, 
we  find  that  R  ^  can  have  no  component  perpen- 
dicular to  the  plane.    Thus  the  four  remaining 

forces 

P,     Q,     -(P  +  Q),    R^, 

are  all  in  one  plane. 


-(P+Q) 


*A 


Fig.  54 


COUPLES  101 

Next,  resolving  parallel  and  jierpentiicular  to  the  line  of  action  of  P,  we 
find  that  both  components  of  R^  vanish,  and  hence  that  72^  =  0.  Thus  the 
original  forces  were  in  eijuilibriiun. 

76.  Clearly  these  methods  of  compounding  forces  can  be 
extended,  so  that  any  number  of  parallel  forces  can  be  compounded 
into  a  single  resultant  force.  AVe  see  at  once,  on  reversing  the 
resultant  and  resolving,  that  the  resultant  is  parallel  to  the  lines 
of  action  of  the  original  forces,  while  its  magnitude  is  equal  to 
their  algebraic  sum. 

This  result  is  of  importance  in  connection  with  the  weights  of 
bodies.  It  shows  that  the  effect  of  gravity  on  any  rigid  body  — 
i.e.  the  resultant  of  the  weights  of  the  individual  particles  of  which 
the  body  is  composed  —  may  be  regarded  as  a  single  force  acting 
vertically  along  a  single  line. 

In  the  next  chapter  it  will  be  shown  that,  whatever  position  the 
rigid  body  is  in,  this  line  always  passes  through  a  detinite  point, 
fixed  relatively  to  the  body,  known  as  its  center  of  gravity. 

77.  Without  assuming  this,  we  can  find  the  line  of  action  in  a 
number  of  simple  cases.  Suppose,  for  mstance,  we  are  dealing  with 
a  uniform  rod.  The  weights  of  two  equal  particles  equidistant  from 
the  center  may  be  compounded  mto  a  single  force  acting  through 
the  middle  point  of  the  rod.  Treating  the  weights  of  all  the  parti- 
cles in  this  manner,  we  find  that  the  weight  of  a  uniform  rod  may 
be  supposed  to  act  at  its  middle  point. 

In  the  same  way  we  can  see  that  the  weight  of  a  circular  disk, 
of  a  circular  ring,  or  of  a  sphere  may  be  supposed  to  act  at  its 
center  ;  the  weight  of  a  cube  or  a  parallelepiped  at  the  intersection 
of  its  diagonals,  and  so  on. 

Couples 

78.  If  we  try  to  compound  two  parallel  forces  which  are  equal 
in  magnitude  but  opposite  in  sign,  we  obtain  as  the  resultant  a 
force  of  zero  amount  of  which  the  line  of  action  is  at  infinity. 
Althoucrh  such  a   force  is  of  zero   amount,  its   effect   cannot  be 


102 


STATICS  OF   RIGID   BODIES 


/^R 


neglected :  its  moment  does  not  vanish,  being  equal  to  the  sum  of 
the  moments  of  the  component  forces.  If,  in  fig.  55,  AA',  BB'  are 
the  parallel  lines  of  action  of  two  opposite  forces  each  equal  to  R, 

and  if  PAB  is  a  Ime  at  right  angles  to 
their  direction,  then  the  sum  of  their 
moments  about  a  line  through  P,  at 
right  angles  to  the  plane  in  which  the 
forces  act, 
~^^  =  11  ■  PB  -R-  PA 

=  Rd, 

where  d  is  the  distance  between  the 
Kne  of  action  of  the  forces.  A  pair  of 
forces,  equal  in  magnitude  and  opposite 
in  direction,  but  not  acting  in  the  same 
line,  is  called  a  couple.  Their  moment  about  any  point  P  in  the 
plane  containing  their  lines  of  action  is  independent  of  the  position 
of  the  point  P,  and  is  spoken  of  as  the  'moment  of  the  couple. 


Ii'> 


Fig. 


Condition  of  Equilibrium 

79.  Since  the  resultantof  a  system  of  forces  in  a  plane  may  be 
either  a  single  force  or  a  couple,  the  condition  for  there  being  no 
resultant  will  be  that  the  resultant  single  force  shall  be  nil,  and 
that  there  shall  be  no  couple.  The  component  of  the  resultant 
force  in  any  direction  vanishes  if  the  sum  of  the  components 
vanishes.  Thus,  in  order  that  the  resultant  force  may  vanish,  it  is 
necessary  that  the  resolved  parts  in  two  different  directions  should 
vanish.  If  this  condition  is  satisfied,  there  can  be  no  resultant 
except  a  couple,  and  since  the  moment  of  a  couple  is  the  same,  no 
matter  about  what  point  the  moment  is  taken,  it  appears  that  there 
can  be  no  couple  if  the  moment  about  any  one  point  is  zero.  Thus, 
as  a  necessary  and  sufficient  condition  of  equilibrium  for  a  system 
of  forces  in  a  plane,  we  have  found  the  following : 

A  system  of  forces  in  a  plane  loill  he  in  equilibrium  if  the  su7n 
of  the  resolved  parts  in  two  directions  each  vanishes,  and  if  the  sum 
of  the  Tnoments  about  any  point  also  vanishes. 


CONDITION  OF   EQUILIBRIUM  103 

We  can  express  the  condition  for  equilihriuin  in  a  different  form  : 

A  system  of  forces  in  a  j^lnne  tvill  he  in  equilihrium  if  the  sums 
of  the  moments  about  any  three  non-collinear  points  are  each  zero. 

For,  if  the  moment  about  any  one  point  is  zero,  the  resultant 
cannot  be  a  couple.  It  must,  therefore,  be  a  single  force.  If  the 
moments  about  each  of  two  points  A,  B  vanish,  the  line  of  action 
of  this  force  must  in  general  be  AB,  but  if  the  moment  about 
some  tliird  point  C,  not  in  the  line  AB,  also  vanishes,  then  the 
force  itself  must  vanish. 

EXAMPLES 

1.  Parallel  forces  of  5,  12,  and  7  pounds  act  at  the  two  ends  and  middle 
point,  respectively,  of  a  line  2  feet  in  lengtli.  Find  the  magnitude  and  line  of 
action  of  their  resultant. 

2.  Find  the  resultant  of  the  forces  in  the  last  question  when  their  magnitudes 
are  respectively  5,  —  12,  and  7  pounds. 

3.  Find  the  resultant  of  three  forces,  each  of  amount  P,  acting  along  the 
sides  of  an  equilateral  triangle,  taken  in  order. 

4.  Prove  that  a  system  of  forces  acting  along  and  represented  by  the  sides 
of  a  plane  polygon  taken  in  order,  is  equivalent  to  a  couple,  whose  moment  is 
represented  by  twice  the  area  of  the  polygon. 

5.  If  the  sums  of  the  moments  of  any  co-planar  forces  about  three  points 
which  are  not  in  a  straight  line  are  equal,  and  not  each  zero,  prove  that  the 
system  is  equivalent  to  a  couple. 

6.  A  uniform  rod  is  of  length  3  feet  and  weight  24  pounds.  Weights  of  16 
and  18  pounds  are  clamped  to  its  two  ends.  Find  at  what  point  the  rod  must  be 
supported  so  as  just  to  balance. 

7.  A  uniform  beam  weighing  20  pounds  is  su.spended  at  its  two  ends, 
and  has  a  weight  of  50  pounds  suspended  from  a  point  distant  7  feet  and 
3  feet  from  the  two  ends.  Find  the  pressures  at  the  points  of  suspension  of 
the  beam. 

8.  A  uniform  rod  of  weight  50  pounds  and  length  18  feet  is  carried  on  the 
shoulders  of  two  men  who  walk  at  distances  of  2  feet  and  3  feet  respectively 
from  the  two  ends.  A  weight  of  50  pounds  is  suspended  from  the  middle  point 
of  the  beam.    Find  the  total  weight  carried  by  each  man. 

9.  A  dumb-bell  weighing  32  pounds  is  formed  of  two  equal  spheres,  each  of 
radius  3  inches,  connected  by  a  bar  of  iron  so  that  the  centers  of  the  spheres 
are  16  inches  apart.  One  of  the  spheres  is  now  removed,  and  the  remainder  of 
the  dumb-bell  is  found  to  weigh  20  pounds.  Find  where  this  remaining  part 
must  be  supported  in  order  that  it  may  just  balance. 


104  STATICS   OF  RIGID   BODIES 

Couples  in  Parallel  Planes 

80.  The  result  of  §  79  shows  that  two  couples  acting  m  the 
same  plane  produce  the  same  effect  if  their  moments  are  equal. 
For,  on  reversing  one  of  them,  all  the  conditions  of  equilibrium 
are  satisfied. 

Thus  we  can  determine  the  effect  of  a  couple  in  any  plane  by 
knowing  its  moment  only.  We  shall  now  show  that  the  actual 
plane  in  which  the  couple  acts  is  immaterial,  the  direction  of  this 
plane  alone  being  of  importance.    In  other  words : 

Couples  of  equal  iiioments  acting  in  23arallel  planes  produce  the 
same  effect. 

To  prove  this,  we  reverse  one  couple  and  show  that  the  two 
couples  are  then  in  equilibrium.    Let  the  first  couple  consist  of 

two  forces  each  equal  to  Pi, 
and  let  a  common  perpen- 
dicular to  their  lines  of 
action  meet  the  latter  in 
points  A,  B.  Let  A'B'  be 
a  line  equal  and  parallel 
to  AB  in  the  plane  of  tlie 
second  couple,  and  let  the 
second  couple  reversed  be 
represented  by  two  forces 
B,  R  at  A'B'..  We  can  regard  this  couple  as  representmg  the  second 
couple  reversed,  for  its  moment  is  equal  and  opposite  to  that  of 
the  second  couple,  while  it  is  acting  in  the  same  plane  as  the 
second  couple. 

We  have  now  to  show  that  the  four  forces  each  equal  to  R 
acting  at  A,  B,  A',  B'  are  in  equilibrium.  By  construction,  ABB' A'  is 
a  parallelogram,  so  that  C,  the  point  of  intersection  of  its  diagonals, 
is  also  the  middle  point  of  each  diagonal. 

The  two  parallel  forces  R,  R  acting  at  A,  B'  may  be  compounded 
into  a  smgle  force  2  R  acting  at  C,  the  middle  point  of  AB',  and 
similarly  the  two  forces  R,  R  at  B,  A'  may  be  compounded  into 


COMPOSITION   OF   COUPLES 


105 


a  force  2  E  Sit  C,  the  middle  point  of  A'B.  These  two  forces  2  E, 
2  E  are  equal  and  act  in  opposite  directions  at  the  same  point  C. 
There  is  therefore  equilibrium,  proving  that  two  couples  are  equiv- 
alent if  they  have  equal  moments  and  if  the  planes  m  which  they 
act  are  parallel. 

81.  The  direction  perpendicular  to  the  plane  in  which  a  couple 
acts  is  called  the  axis  of  a  couple.     Thus : 

Ttvo  couples  ■haviiuj  the  same  axis  and  the  same  moment  are 
equivalent. 


Couples  compounded  according   to  the  Parallelograim  Law 

82.  A  couple,  as  we  have  just  seen,  is  determined  by  a  quantity 
(its  moment)  and  a  direction  (its  axis).  Thus  it  may  be  fully 
represented  by  a  straight  line,  the  direction  of  tliis  line  being  that 
of  the  axis,  and  the 
length  representing, 
on  any  scale  we 
please,  the  magni- 
tude of  the  moment 
of  the  couple. 

We  shall  now 
show  that  couples 
can  be  compounded 
according  to  the 
parallelogram  law. 

Theorem.  Two  couples,  represented  in  magnitude  and  direction 
hy  two  lines  AB,  AC,  tvill  have  as  resultant  a  couple  represented  in 
magnitude  and  direction  hy  AD,  the  diagonal  of  the  parallelogram 
of  which  AB,  AC  are  edges. 

Let  AB,  AC  be  Imes  representing  by  their  direction  and  magni- 
tude the  axes  and  moments  of  two  couples.  Let  SAS'  be  a  line 
perpendicular  to  the  plane  ABC,  A  being  its  middle  pomt.  Let 
us  draw  planes  through  S,  S'  parallel  to  the  plane  ABC,  and  let 
the  couple  AB  he  replaced  by  two  forces  PS,  F'S'  in  these  two 


Fig.  57 


lOG  STATICS   OF   RIGID   BODIES 

planes,  the  lines  FS,  F'S'  being  both  perpendicular  to  AB.  In  the 
same  way  let  the  couple  ^C  be  replaced  by  two  forces  QS,  Q'S' 
in  these  same  two  planes. 

The  two  couples  have  now  been  replaced  by  the  four  forces 
PS,  QS,  F'S',  Q'S'. 

Let  us  complete  the  parallelograms  FSQE,  BACD,  F'S'Q'Ii'. 
Obviously  these  parallelograms  are  all  similar  to  one  another,  and 
corresponding  lines  in  the  first  and  second  parallelograms  are  at 
right  angles  to  one  another.  Thus  a  couple  represented  by  AD 
may  be  replaced  by  forces  RS,  E'S'.  But  these  two  forces  are 
exactly  equivalent  to  the  four  forces  FS,  QS,  F'S',  Q'S'  to  which, 
as  we  have  seen,  the  couples  AB,  AC  may  be  reduced,  and  this 
proves  the  theorem. 

FoKCES  IN  Space 

83.  When  the  forces  acting  on  a  body  are  not  all  in  one  plane, 
their  resultant  will  not  in  general  be  a  single  force. 

Theorem.  Any  system  of  forces  acting  on  a  rigid  hody  can  he  re- 
placed hy  a  force  acting  at  an  arbitrarily  chosen  point,  and  a  couple. 
Let  G  be  the  chosen  point,  and  let  R  be  any  force  of  which  the 
line  of  action  does  not  pass  through  G.    At  G  let  us  introduce  two 
equal  and  opposite  forces,  each  equal  to  R 
and  parallel  to  the  line  of  action  of  R.    By 
combining  one  of   these   forces  with  the 
original  force  R,  we  get  a  couple,  so  that 
the  original  force  R  can  be  replaced  by  a 
force    parallel    and   equal   to   the   original 
force  but  actuig  at  G,  and  a  couple. 

Treating  all  the  forces  of  the  system  in  this  way,  we  find  that 
the  origmal  system  of  forces  may  be  replaced  by 

(a)  a  number  of  forces  actmg  at  the  chosen  point  G; 

(b)  a  number  of  couples. 

The  forces  acting  at  G  can  be  combined  into  a  single  force  at  G, 
and  the  couples  into  a  single  couple,  proving  the  result. 


FORCES  IN   SPACE 


107 


84.  Theorem.  Any  system  of  forces  acting  on  a  rigid  hody  can 
he  replaced  hy  a  force  and  a  couple  of  ivhicJi  the  axis  is  parallel  to 
the  line  of  action  of  the  force. 

By  the  theorem  just  proved,  tlie  system  may  first  be  replaced 
by  a  force  acting  at  any  point  0,  and  a  coiqile.  Let  the  force  be 
of  amount  R,  having  OF  as  its  line  of  action,  and  let  the  couple 
be  of  moment  G,  having  OQ  for  its  axis.  If  the  angle  PO^  is 
denoted  by  Q,  we  may  resolve  tlie  couple  into  two  couples: 

{a)  a  couple  of  moment  G  cos  Q, 
having  OP  for  axis; 

(h)  a  couple  of  moment  G  sin  6, 
having  its  axis  perpendicular  to  OP. 

The  second  of  these  couples  may 
be  replaced  by  any  two  forces  pro- 
vided these  are  chosen  so  as  to  be 
equivalent  to  the  couple.  Let  us 
choose  the  first  force  to  be  a  force 
—  R  acting  along  OP,  i.e.  the  force 
which  will  exactly  neutralize  the 
force  R  which  we  already  have  acting  along  OP.  The  second 
force  of  the  couple  must  then  be  a  force  R  acting  along  a  Hne 
parallel  to  OP  but  at  a  distance  from  it  equal  to  G  sin  dJR. 

The  system  has  now  been  replaced  by 

(a)  forces  -\-  R,  —  R  acting  along  OP ; 

(h)  a  force  R  parallel  to  0P\ 

(c)  a  couple  G  cos  Q  of  which  the  axis  is  parallel  to  OP. 

The  two  forces  {(()  neutralize  and  we  are  left  with  a  force  R  and 
a  couple  G  cos  6  of  which  the  axis  is  parallel  to  the  line  of  action 
of  the  force.    This  proves  the  theorem. 

The  line  of  action  of  the  force,  which  is  now  also  the  axis  of 
the  couple,  is  called  the  central  axis  of  the  system  of  forces. 
A  system  of  forces  is  most  simply  specified  by  a  knowledge  of  the 
magnitude  of  the  force  and  couple,  and  of  the  position  and  direction 
of  the  central  axis.    Such  a  system  is  called  a  "  wrench." 


1U8 


STATICS  OF  KIGID  BODIES 


Fig.  60 


ILLUSTRATIVE   EXAMPLES 

1.  Two  equal  uniform  planks  are  hinged  at  one  end,  and  stand  with  their  free 
ends  on  a  smooth  horizontal  plane,  being  prevented  from  slipping  hij  a  rope  which 
is  tied  to  each  plank  at  the  same  height  up.  Find  the  tension  in  this  rope  and 
the  action  at  the  hinge. 

In  the  figure  let  AB,  AC  represent  the  two  planks,  hinged  at  A,  and  let 
PQ  be  the  rope.    The  forces  acting  on  the  plank  AB  will  consist  ot 

(a)  the  action  at  the  hinge  A  ;' 

(b)  the  tension  of  the  rope  acting  along  PQ ; 

(c)  the  reaction  at  the  foot  B ; 

(d)  the  weight. 

Of  the.se  four  forces,  (a)  and  (6)  are  the  forces  which 
it  is  required  to  find.     Force  (c)  also  is  at  present 
unknown.    Force  (d),  as  explained  in  §  77,  can  be 
regarded  as  a  single  force  W,  the  total  weight  of  the 
C  plank,  and  since  we  are  told  that  the  plank  is  uniform, 

this  must  be  supposed  to  act  through  its  middle  point. 
There  is  a  simple  way  of  finding  force  (c),  the  reaction  at  B.  Since  we  are 
told  that  the  contact  at  B  is  smooth,  the  direction  of  the  reaction  must  be 
vertically  upward.s.  Let  its  amount  be  R.  From  symmetry,  there  must  be 
an  exactly  similar  reaction  at  the  foot  C  of  the  .second  plank.  Now  consider 
the  equilibrium  of  the  whole  system  which  consists  of  the  two  planks  and  the 
rope.    The  only  external  forces  which  act  on  this  system  consist  of 

(a)  the  weight ; 

(6)  the  two  reactions  at  B  and  C. 

If  we  resolve  vertically,  we  obtain,  .since  this 
system  is  in  equilibrium, 

2W-2R  =  0, 
so  that  R  —  W ;  each  reaction  is  just  equal  to  the 
weight  of  one  plank,  as  we  might  have  anticipated. 

Of  the  four  forces  acting  on  the  plank  AB, 
the  last  two  are  now  known,  while  the  first  two 
remain  unknown.  If  we  take  moments  about  A, 
we  shall  get  an  equation  between  forces  (b),  (c), 
and  (d),  and  this  will  enable  us  to  find  the  un- 
known force  (b),  the  tension. 

If  we  denote  the  tension  by  T,  and  the  angle 
BA  C  hj  2  6,  the  equation  obtained  on  taking 
moments  about  A  is 

R-  AB  sine  -  W-^AB  sin  6  -  T-APcosd 
so  that,  remembering  that  R  =  W,  vfe  have 

„      I  AB 


Fig.  G1 


0, 


2  AP 


tan  6. 


ILLUSTRATIVE   EXAMPLES 


109 


Also  on  resolving  horizontally  and  verticaliy,  it  is  evident  that  the  action 
at  A  must  consist  of  a  horizontal  force  of  amount  T  and  of  direction  opposite 
to  that  of  T. 

2.  A  ring  {e.g.  a  dinner  nnpkln  ring)  stands  on  a  table,  and  a  gradually 
increasing  pressure  is  applied  by  a  finger  to  one  point  on  the  ring.  Having  given 
the  coefficients  of  friction  at  the  two  contacts,  examine  how  equilibrium  will  first 
he  broken. 

Let  A  be  the  point  of  contact  of  the  ring  and  table,  and  let  B  be  the  point 
of  contact  of  the  ring  and  finger.  Let  e,  e'  be  the  angles  of  friction  at  A  and  B 
respectively.    Let  the  line  BA  make  an  angle  a  with  the  vertical. 


The  forces  applied  to  the  ring  from  outside  are 
(a)  the  reaction  at  A  ; 
(6)  the  reaction  at  B ; 
(c)  the  weight  of  the  ring. 

Regarding  the  latter  as  a  single  force  W  acting  along  the  vertical  diameter 
CA  of  the  ring,  we  see  that  so  long  as  the  ring  remains  at  rest  it  is  in  equilibrium 
under  the  action  of  three  forces. 

Hence,  by  the  theorem  of  §  52,  the  lines  of  action  of  the  three  forces  must 
meet  in  a  point. 

The  line  of  action  of  the  weight  is  already  known  to  be  the  vertical  CA, 
and  the  line  of  action  of  the  reaction  at  A  must  pass  through  A.    Hence  either 

(or)  the  point  in  which  the  three  lines  of  action  meet  must  be  A  ;  or 

(/3)  the  reaction  at  A  must  act  along  CA,  so  that  the  point  in  which  the 
three  lines  of  action  meet  will  be  some  point  in  CA,  other  than  A. 

The  second  alternative  may  be  dismissed  at  once.  For  if  the  reaction  at  A 
acts  along  CA,  this  and  the  weight  may  be  combined  into  a  single  force,  and 
there  must  now  be  equilibrium  under  this  force  and  the  reaction  at  B.  This 
requires  that  each  force  should  vanish,  i.e.  there  must  be  no  pressure  at  B, 
and  the  reaction  at  A  must  be  just,  equal  to  the  weight  of  the  ring.    This 


no 


STATICS   OF  Kit  J  ID   JJUDIES 


obviously  gives  a  state  of  equilibrium  —  the  ring  is  standing  at  rest  on  the  table, 
acted  on  solely  by  its  weight  —  but  this  state  of  equilibrium  is  not  the  one  with 
which  we  are  concerned  in  the  present  problem. 

Let  us  now  consider  the  meaning  of  alternative  (a).  If  the  three  lines  of 
action  meet  in  A,  the  reaction  at  B  must  act  along  BA^  and  this  must  be  true 
no  matter  how  great  the  pressure  at  B.  Hence  the  reaction  at  B  will  always 
make  an  angle  a  with  the  normal. 

If  a  is  less  than  e',  the  angle  of  friction  at  B,  this  will  be  a  possible  line  of 
action  for  the  reaction,  and  no  slipping  can  take  place  at  B,  no  matter  how 
great  the  pressure  applied  at  B  may  be. 

On  the  other  hand,  if  a  is  greater  than  e',  equilibrium  is  impossible,  no  matter 
how  small  the  pressure  applied  at  B  may  be.  Thus,  when  there  is  equilibrium, 
the  pressure  at  B  must  vanish,  and  we  are  led  to  the  same 
state  of  equilibrium  as  was  reached  from  case  (/3).  As  soon 
as  the  pressure  at  B  becomes  appreciable,  equilibrium  is 
obviously  broken  by  slipping  taking  place  at  B,  since  for 
equilibrium  to  be  maintained  at  B,  the  reaction  would 
have  to  act  at  an  angle  greater  than  the  actual  angle  of 
friction. 

Thus  the  solution  resolves  itself  into  two  different  cases : 
Case  I.    If  a  >  e',  as  soon  as  pressure  is  applied  at  jB, 
motion  takes  place.    The  ring  slips  at  B,  and  consequently 
rolls  at  A. 

Case  II.  If  a  <  e',  we  have  seen  that  no  matter  .how 
great  a  pressure  is  applied  at  B,  there  can  never  be 
slipping  at  B.  It  remains  to  examine  whether  there  can 
be  slipping  at  A. 
To  settle  this  question,  we  have  to  determine  whether  the  reaction  at  A  can 
ever  be  made  to  act  at  an  angle  with  the  vertical  which  is  as  great  as  the  angle 
of  friction  at  A,  namely  e.  Now  the  ring  is  acted  on  by  three  forces,  the 
reactions  at  A  and  B,  say  Ra  and  Rb,  and  its  weight  IF.  The  lines  of  action 
of  these  forces  meet  in  the  point  A^  and  from  Lami's  theorem  we  can  connect 
the  magnitudes  of  the  forces  with  the  angles  between  them. 

The  lines  of  action  of  the  three  forces  are  represented  in  fig.  63.  The  angle 
between  W  and  Eg  is,  as  we  have  seen,  always  equal  to  a.  Let  the  angle 
between  Ra  and  the  vertical  be  suppo.sed  to  be  e.    Then,  by  Lami's  theorem, 

W 


Fig.  G.3 


Ra 

sin  a 


Rb 

sin^ 


sin  [a  —  d) 


The  value  of  Ra  is  not  given,  but  on  ecjuating  the  last  two  fractions  we  have 

W  _  sin  (a  — 
Rjg  sin  6 

so  that 


sin  a  cot  0  —  cos  a, 


W\ 
cot  d  =  I  cos  a  H )  cosec  a. 

Rb) 


EXAMPLES  111 

This  equation  enables  us  to  trace  the  clianges  in  the  valuo  of  the  angle  d  as 
Rb  is  gradually  increased.  We  tind  that  when  Rjj  =  0  the  value  of  0  is  6  =  0, 
and  that  as  En  increases  0  increases  continually,  but  never  exceeds  the  value 
6  =  a,  which  is  reached  when  Rjj  =  co. 

If  e  <  a,  the  value  of  6  will  pass  throui^h  the  value  e  when  Rn  reaches  a 
certain  value,  namely 

sine 

liji  = 1 

T^sin(a  —  e) 

and  slipping  at  A  will  take  place  at  this  point. 

If  e>a,  the  value  of  6  will  never  reach  the  value  e,  so  that  slipping  at  A  can 
never  occur.  Thus  equilibrium  is  never  broken,  and  the  harder  we  press  at  B 
the  more  firmly  the  ring  is  held  between  the  finger  and  the  table. 

We  can  now  summarize  the  results  which  have  been  obtained,  as  follows : 

If  a  >  e',  the  ring  rolls  along  the  table  as  soon  as  we  begin  to  press  at  B. 
If  a:  <  e',  there  are  two  cases  : 

(a)  a>e,  the  ring  will  slip  at  A  as  soon  as  sufficient  pressure  is  applied  ; 
(6)  a:  <  e,  the  ring  cannot  be  made  to  move  under  any  amount  of  pressure. 

To  make  the  ring  shoot  out  from  under  the  finger  by  slipping  at  A  (in  wdiich 
case  it  returns  to  the  hand,  as  in  the  well-known  trick),  it  is  necessary  to  press 
at  a  point  on  the  ring  at  which  a  is  greater  than  e,  while  being  less  than  e'. 
We  notice  that  if  e  is  greater  than  e',  it  is  impossible  to  project  the  ring  in  this 
way ;  this  can  only  be  done  if  the  contact  with  the  finger  is  rougher  than  the 
contact  with  the  table. 

GENERAL  EXAMPLES 

1.  A  pair  of  steps  is  formed  by  two  uniform  ladders  each  of  length  12 
feet  and  weight  20  pounds,  jointed  at  the  top,  and  having  their  points  at 
distances  5  feet  from  the  ground  connected  by  a  rope.  The  steps  stand  on 
a  smooth  horizontal  plane,  and  a  man  of  weight  160  jaounds  ascends  to  a 
height  of  9  feet  on  one  side.     Find  the  tension  in  the  rope. 

2.  A  heavy  uniform  rod  is  supported  by  two  strings  of  lengths  o,  b. 
The  upper  ends  of  the  strings  are  tied  to  the  same  point,  the  lower  ends 
being  tied  to  the  two  ends  of  the  rod.  Show  that  the  tensions  of  the  strings 
are  pi'oportional  to  a  and  b  respectively. 

3.  Two  small  fixed  pegs  are  in  a  line  inclined  at  an  angle  0  to  the 
horizon.  A  rough  thin  rod  passes  under  the  lower  and  rests  on  the  higher, 
this  latter  being  lower  than  the  center  of  gravity  of  the  rod.  The  distances 
of  the  center  of  gravity  from  the  two  pegs  are  a  and  b  respectively,  and  the 
coefficient  of  friction  is  fi.     Show  that  if  the  rod  is  on  the  point  of  motion, 

H  = tan  0. 

b  +  a 


112  STATICS  OF  EIGII)   BODIES 

4.  Two  heavy  uniform  rods  have  their  ends  connected  by  two  light 
strings,  and  the  whole  system  is  suspended  by  the  middle  point  of  one  rod. 
Prove  that  in  equilibrium  either  the  rods  or  the  strings  are  parallel. 

5.  Two  rods  AB,  CD  lying  on  a  smooth  table  are  connected  by  stretched 
strings  AC,  BD.  If  the  system  is  kept  in  ecpiilibriiun,  by  forces  acting  at 
the  middle  points  of  the  rods,  prove  that 

(r/)  the  rods  must  be  parallel; 

(J))  the  tensions  must  be  proportional  to  the  strings. 

6.  A  BCD  is  a  parallelogram  and  E  is  the  intersection  of  the  diagonals 
AC,  BD.  Show  that  parallel  forces  7,  5,  16,  4  at  J.,  5,  C,  Z)  respectively 
are  equivalent  to  other  parallel  forces,  8  at  the  middle  point  of  CD,  10  at 
the  middle  point  of  BC,  and  14  at  E. 

7.  A  solid  cube  is  placed  on  a  rough  inclined  plane  of  angle  a  with  two 
edges  of  its  base  along  lines  of  greatest  slope.  The  angle  of  friction  is  e. 
Prove  that  if  a  >  45°  it  will  at  once  topple  over,  while  if  e  <  a  <  45°  it  will 
slide  down  the  plane.  If  a  is  less  than  either  e  or  45°,  find  the  friction 
brought  into  action. 

8.  A  uniform  rod  of  length  2  /  and  weight  W  rests  over  a  smooth  peg 
at  distance  /*  (<  I)  from  a  smooth  vertical  wall  at  an  angle  d  with  the  hori- 
zontal, its  lower  end  pressing  against  the  wall,  and  its  upper  being  held  by  a 
vertical  string.    Find  the  tension  of  the  string  and  show  that  it  vanishes  if 


e  =  COS-l  -»  ;  I  - 

\\l 

9.  Two  equal  uniform  spheres,  each  of  weight  W  and  radius  a,  rest  in 
a  smooth  hemisjiherical  bowl  of  radius  b.  Find  the  pressure  between  the 
two  spheres  and  also  the  pressm-e  of  each  on  the  bowl. 

10.  A  uniform  rod  rests  with  its  two  ends  on  smooth  inclined  jilanes, 
inclined  to  the  horizontal  at  angles  a  and  /3.  Find  the  inclination  of  the 
rod  to  the  horizontal. 

11.  In  the  last  question  a  weight  equal  to  that  of  the  rod  is  clamped  to  it. 
At  what  point  must  it  be  clamped  in  order  that  the  rod  may  rest  horizontally? 

12.  A  uniform  circular  ring  of  weight  W  has  a  bead  of  weight  iv  fixed 

on  it  and  hangs  on  a  rough  peg.    Show  that  if  sin  e> ,  then  the  ring 

W  +  w 
can  rest  without  slipping,  whatever  point  of  it  rests  on  the  peg,  e  being 
the  angle  of  friction. 

13.  A  pentagon  ABCDE,  formed  of  equal  uniform  heavy  rods  connected 
by  smooth  joints  at  their  ends,  is  supported  symmetrically  in  a  vei'tical 
plane  with  A  uppermost,  and  AB,  AE  in  contact  with  two  smooth  pegs 
in  the  same  horizontal  line.  Prove  that  if  the  pentagon  is  regular,  the 
pegs  must  divide  AB  and  AE  each  in  the  ratio 

1  +  sin  Jq  TT  :  3  sin  -^^  ir. 


EXAMPLES  113 

14.  A  uniform  beam  of  leugtli  /  leans  against  the  horizontal  rim  of  a 
hemispherical  bowl  of  radius  a,  with  its  lower  end  resting  upon  the  smooth 
concave  surface.     Find  its  inclination  to  the  vertical. 

15.  A  bowl  in  the  shape  of  a  paraboloid  of  revolution  is  placed  with  its 
axis  vertical.  A  uniform  rod  rests  on  a  peg  at  the  focus  and  has  its  lower 
end  resting  on  the  inner  surface.  Both  contacts  are  perfectly  smooth. 
Find  the  inclination  of  the  rod  to  the  vertical. 

16.  A  miiform  beam  of  weight  W  rests  against  a  vertical  wall  and  a 
horizontal  plane  with  which  it  makes  the  angle  a.  Both  contacts  are 
perfectly  smooth.  The  lower  end  of  the  beam  is  attached  by  a  string  to 
the  foot  of  the  wall.     Find  the  tension  of  the  string. 

17.  One  end  of  a  straight  uniform  heavy  rod  rests  on  a  rough  liorizontal 
plane,  the  other  end  being  connected  with  a  fixed  point  by  a  string.  If 
6,  0,  i//  be  the  inclinations  of  the  string,  the  rod,  and  the  total  reaction  of 
the  horizontal  plane  respectively  to  the  vertical,  show  that 

cot  ^  ±  2  cot  (/>  -  cot  i/-  =  0. 

18.  Two  uniform  rods  AB,  BC  of  the  same  material  but  of  different 
lengths  are  jointed  freely  at  B  and  fixed  to  a  vertical  wall  at  A  and  C.  Show 
that  the  direction  of  the  reaction  at  B  bisects  the  angle  ABC. 

19.  A  uniform  regular-hexagonal  board  ABCDEF  of  given  weight  W 
is  supported  in  a  horizontal  position  on  tliree  pegs,  placed  at  the  corners 
A ,  B  and  the  middle  point  of  DE.    Find  the  pressures  on  the  pegs. 

20.  Two  spheres  of  radii  a,  h  and  weights  TF,  W  respectively  are 
suspended  freely  by  strings  of  lengths  I,  I'  respectively  from  the  same  hook 
in  the  ceiling.    If  l'>l  -\-2a,  show  that  the  angle  which  the  first  string 

makes  with  the  vertical  is 

.      ,  TFrt 

sin~ ' • 

{W  +  W'){a  +  I) 

21.  A  uniform  rod  hangs  by  two  strings  of  lengths  /.  /'  fastened  to  its 
ends  and  to  two  hooks  in  the  same  horizontal  line  at  the  distance  a.  If 
the  strings  cross  one  another  and  make  the  respective  angles  a,  a',  with 
the  horizontal,  show  that  when  the  rod  is  in  equilibrium 

sin  (a  +  a')  (r  cos  a'  —  I  cos  tr)  =  a  sin  (a  —  «'). 

22.  A  uniform  planlc  of  length  2  b  rests  with  one  end  on  a  rough  hori- 
zontal plane,  touches  a  smooth  fixed  cylinder  of  radius  a  lying  on  the  plane, 
and  makes  an  angle  2  a  with  the  plane,  the  angle  of  friction  being  e. 
Show  that  equilibrium  is  possible  if 

a  sin  e>b  tan  a  cos  2  a  sin  (2  a  +  e) . 

23.  Two  equal  and  similar  isosceles  wedges,  each  of  weight  TF  and 
vertical  angle  2  a,  are  placed  side  by  side  with  their  bases  on  a  rough 


114  STATICS  OF  RIGID   BODIES 

horizontal  table  so  as  to  be  just  in  contact  along  an  edge.  A  smooth 
si:)here  of  weight  ic  and  radius  r  is  supported  between  them,  being  in  con- 
tact with  a  face  of  each.     Prove  tliat  for  ecjuilibrium  it  is  necessary  that 

ic  cot  a  r,       •        J.  /i    ,  ^'  \ 

u  > ,     r  <2  a  sin  ex  tan  a  (  1  H ) , 

2  W  +  IV  \  in  / 

where  m  denotes  the  coefficient  of  friction  and  2  a  is  the  length  of  either 

base. 

24.  A  seesaw  consists  of  a  jilank  of  weight  w  laid  across  a  fixed  rough 
log  whose  shape  is  that  of  a  horizontal  circular  cylinder.  The  inclination 
to  the  horizontal  at  which  it  balances  is  increased  to  a  when  loads  W,  W 
are  placed  at  the  lower  and  higher  ends  respectively ;  and  the  inclination 
is  reduced  to  ^  when  these  loads  are  interchanged.  Show  that  the  inclina- 
tion of  the  plane  when  unloaded  is 

w'jW  +  W'+  w)  (W'a  -  W^^ 

iv(W  +  W'  -  iv')(\V  -W)    ' 
w'  being  the  weight  which,  placed  at  the  liigher  end,  would  balance  the 
plank  horizontally. 

25.  A  chain  is  formed  of  2  n  exactly  similar  links,  the  contacts  between 

consecutive  links  being  perfectly  smooth.     The  two  end  links  can  slide  on 

a  horizontal  wire,  the  contact  here  being  rough,  coefficient  of  friction  /u. 

Show  that  in  the  limiting  position  of  equilibrium,  the  inclination  of  either 

of  the  upper  links  to  the  vertical  is 

2  ??M 

tan- 1 

2/1-1 

26.  Two  equal  circular  disks  of  radius  r,  with  smooth  edges,  are  placed 
on  their  flat  sides  in  the  corner  between  two  smooth  vertical  planes  inclined 
at  an  angle  2  a,  and  touch  each  other  in  the  line  bisecting  the  angle. 
Prove  that  the  smallest  disk  which  can  be  pressed  between  them  without 
causing  them  to  separate  is  one  of  radius  r(sec  a  —  1). 

27.  How  is  the  result  of  the  last  question  modified  if  all  the  contacts 
are  rough,  the  angle  of  friction  at  each  being  e? 

28.  Two  uniform  ladders  are  jointed  at  one  end  and  stand  with  their 
other  ends  on  a  rough  horizontal  plane.  A  man  whose  weight  is  equal  to 
that  of  one  of  the  ladders  ascends  one  of  them.  Prove  that  the  other  will 
slip  first. 

If  it  begins  to  slip  when  he  has  ascended  a  distance  x,  prove  that  the 

coefficient  of  friction  is  — —  tan  ex,  a  being  the  length  of  each  ladder 

'2a  +  X 
and  a  the  angle  each  makes  with  the  vertical. 

29.  A  weightless  ladder  rests  against  a  smooth  cube  of  weight  W, 
standing  on  smooth  ground,  with  the  foot  of  the  ladder  tied  to  the  middle 


EXAMPLES  115 

point  of  one  of  the  lowest  edges  of  the  cube ;  a  man  of  weight  rr  ascends 
the  ladder.  Pi-ov^e  that,  if  the  ladder  ])rojects  above  the  top  of  the  cube, 
the  cube  will  tilt  before  he  reaches  the  toj)  of  the  cube  unless 

W>2  tv  cos  (^(siu  a  —  cos  a), 
where  a  is  the  inclination  of  the  ladder  to  the  horizontal. 

30.  Four  equal  spheres  rest  in  contact  at  the  bottom  of  a  smooth 
spherical  bowl,  their  centers  being  in  a  horizontal  plane.  Show  that  if 
another  equal  sphere  be  placed  upon  them  the  lower  spheres  will  scjiarate 
if  the  radius  of  the  bowl  be  greater  than  (2  vl8  +  1)  times  tlie  radius  of 
a  sphere. 

31.  Three  equal  spheres  rest  in  contact  on  a  smooth  horizontal  plane, 
so  that  their  centers  form  an  equilateral  triangle,  and  are  bound  together 
by  a  fine  string  passing  around  them  on  the  level  with  their  centers.  If 
another  equal  sphere  be  placed  symmetrically  on  them,  show  that  the 

tension  of  the  string  is  increased  by  — —  to,  where  w  is  the  weight  of  the 
upper  sphere. 

32.  A  right  circular  cone  of  vertical  angle  2  a  rests  with  its  base  on  a 
rough  horizontal  plane.  A  string  is  attached  to  the  vertex,  and  is  jiulled 
in  a  horizontal  direction  with  a  gradually  increasing  force.  Find  in  what 
way  equilibrium  will  first  be  broken. 

33.  A  heavy  particle  is  placed  on  a  rough  inclined  jilane  wliose  inclina- 
tion is  exactly  equal  to  the  angle  of  friction.  A  thread  is  attached  to  tlie 
particle  and  is  passed  through  a  hole  in  the  plane  which  is  lower  than  the 
particle,  but  is  not  in  the  line  of  greatest  slope  through  it.  Sjiow  that  if 
the  thread  be  gradually  drawn  through  the  hole,  the  particle  will  describe 
a  straight  line  and  a  semicircle  in  succession. 

34.  A  uniform  cubical  block  of  weight  W  rests  with  one  edge  horizontal 
on  a  rough  inclined  plane,  and  against  the  block  rests  a  rough  sphere  of 
weight  W  of  radius  less  than  an  edge  of  the  cube.  The  inclination  of  the 
plane  is  gradually  increased.  Examine  the  different  ways  in  which  equi- 
librium may  be  broken,  and  determine  which  will  actually  occur  in  a 
given  case. 

35.  A  rough  uniform  rod  is  placed  on  a  liori/.ontal  plane  and  is  acted 
on,  at  one  of  the  points  of  trisection  of  its  length,  by  a  horizontal  force  in 
a  direction  perpendicular  to  its  length.  Find  about  what  point  the  rod 
will  begin  to  turn. 

36.  A  heavy  bar  AB  is  suspended  by  two  equal  sti'ings  of  length  /, 
which  are  originally  parallel.  Find  the  couple  which  must  be  applied  to 
the  bar  to  keep  it  at  rest  after  it  has  been  twisted  tlirough  an  angle  0  in 
the  horizontal  plane. 


IIG  STATICS   OF  RKilD   BODIES 

37.  The  line  of  hinges  of  a  door  is  inclined  at  an  angle  a  to  the  vertical. 
Show  that  tlie  conple  recjuired  to  keep  it  in  a  position  inclined  at  an  angle 
(i  to  tliat  of  eqnilibriuni  is  proportional  to  sin  a  sin /3. 

38.  Show  that  any  system  of  forces  acting  on  a  rigid  body  can  be 
rednced  to  two  equal  forces  equally  inclined  to  the  central  axis. 

39.  Prove  that  the  central  axis  of  two  forces  P  and  Q  intersects  the 
shortest  distance  c  between  their  lines  of  action  and  divides  it  in  the  ratio 

Q(Q  +  P  cose):  P(P  +  Q  cos  d) , 

6  being  the  angle  between  their  directions.     Also  prove  that  the  nionicnt 

of  the  principal  conjile  is 

cPQ  sin  9 


VP2  +  Q2  +  2  pQ  cos  d 

40.  Prove  that  the  axis  of  the  resultant  of  two  given  wrenches  (/tj,  //j) 
and  (/'.,,  //.,),  the  axes  of  which  are  inclined  to  each  other  at  an  angle  6, 
intersects  the  shortest  distance,  2  c,  between  their  axes  at  a  point  the  dis- 
tance of  which  from  the  middle  point  is 

{R^  -  Rj)  c  +  {H^R^  -  H^R^)  sin  e 
R^-  +  R.f  +  2 R^R^  cose 


CHAPTEE  YI 


CENTER   OF   GRAVITY 

85.  As  we  have  seen,  the  action  of  gravity  on  a  system  of 
masses  may  be  represented  by  a  system  of  parallel  forces,  these 
forces  consisting  of  a  force  acting  on  each  particle  equal  to  the 
weight  of  the  particle,  its  direction  being  vertically  dowiiwards. 
By  the  rules  explained  in  the  last  chapter,  these  forces  may  be 
compounded  hito  a  single  force.  The  magnitude  of  this  force  is 
the  sum  of  all  the  component 
forces,  and  is  therefore  the  total 
weight  of  the  body,  while  the 
direction  of  the  force,  being  par- 
allel to  the  component  forces, 
is  itself  vertically  downwards. 
The  problem  discussed  in  the 
present  chapter  is  that  of  de- 
termming  the  position  of  the 
line  of  action  of  this  force. 

86.  Let  the  particles  be  of 
masses  m^,  m^,  ■■■.  Let  rectangular  axes  be  taken,  the  axis  of  z 
being  vertical,  and  let  the  coordinates  of  the  first  particle  be 
x^,  y^,  z^,  the  coordinates  of  the  second  be  x„,  y.,,  z.,,  and  so  t)n. 

The  weight  of  the  first  particle  is  7)i^g,  and  its  line  of  action 
cuts  the  plane  Oxy  in  a  point  of  which  the  coordinates  are 
x^,  y^,  0.  Hence  the  moment  of  the  force  about  the  axis  Oy  is 
m^gx^.  Let  the  line  of  action  of  the  resultant  cut  the  plane  Oxy 
in  the  point  1\  y,  0.  Then  the  moment  of  the  resultant  about  the 
axis  Oy  is  (^yiii\(jx,  where  V^/^j  is  the  sum  of  the  masses  of  all 
the  particles. 

117 


Fig.  64 


118  CENTER   OF  CxRAVITY 

Since  the  moment  of  the  resultant  is  equal  to  the  sum  of  the 
moments  of  the  separate  forces,  we  must  have 


^v> 


so  that  X  = 


Similarly 


These  equations  determine  the  coordinates  x,  y  of  the  point  in 
which  the  line  of  action  of  the  resultant  meets  the  plane  Oxy. 

We  have,  however,  seen  that  the  coordinates  of  the  centroid  of 
masses  m^  at  x^,  y^,  z^,  m.,  at  x^,  y„,  z.-^,  etc.,  are 

^m^x^  '^m^y^  ^m^z^ 


X'^^i  X"^i  2'"=^ 

so  that  the  pomt  in  wliich  a  vertical  through  the  centroid  will 
meet  the  plane  Oxy  must  he 

^m^x^       ^m,y^ 

i.e.  the  point  must  he  the  point  x,  y,  0  m  which  the  line  of  action 
of  the  resultant  force  meets  the  plane  Oxy.     Thus 

The  line  of  action  of  the  resultant  force  of  gravity  is  the  ver- 
tical line  through  the  centroid  of  the  particles. 

For  this  reason  the  centroid  of  a  number  of  points,  weighted 
according  to  the  masses  of  the  particles  which  occupy  these  points, 
is  called  the  center  of  gravity  of  the  particles.  The  effect  of 
gravity  acting  on  a  rigid  body  is,  as  we  have  now  seen,  repre- 
sented by  a  single  force  acting  vertically  downwards  through  the 
center  of  gravity  of  the  body,  the  amount  of  the  force  being  equal 
to  the  total  weight  of  the  body.  The  action  of  gravity  is,  accord- 
ingly, the  same  as  if  the  whole  mass  of  the  body  were  concentrated 
in  a  single  particle  placed  at  the  center  of  gravity. 


SYSTEM  OF  MASSES  110 

87.  It  is  clear  that  if  we  suspend  a  rigid  body  or  system  of 
bodies  by  a  string,  the  center  of  gravity  must  be  vertically  below 
the  string.  For  all  the  forces  actmg  on  the  system  reduce  to  two, 
—  the  tension  of  the  string  and  the  weight  acting  at  the  center 
of  gravity,  —  and  in  equilibrium  these  two  must  act  along  the 
same  line. 

In  the  same  way  it  will  be  seen  that  if  a  body  is  phaced  on  a 
point  in  such  a  way  as  to  balance  m  etjuilibrium  on  this  point, 
then  the  center  of  gravity  must  be  vertically  above  the  point. 

88.  A  few  simple  instances  of  the  position  of  the  center  of 
gravity  have  been  mentioned  m  §  77.    These  were  as  follows : 

(a)  the  center  of  gravity  of  a  uniform  rod  is  at  its  middle  point ; 

(b)  the  center  of  gravity  of  a  uniform "  circular  disk,  circular 
ring,  or  sphere  is  at  the  center; 

(c)  the  center  of  gravity  of  a  cube  or  parallelepiped  is  at  the 
center  (i.e.  the  intersection  of  the  diagonals). 

89.  It  is  easy  to  find  the  center  of  gravity  of  a  system  of  bodies 
wdien  the  center  of  gravity  of  each  of  the  component  parts  is  kno^^^l. 
For,  regarding  the  weight  of  each  of  the  bodies  as  a  single  force 
acting  through  its  center  of  gravity,  we  have  a  number  of  parallel 
forces  in  action,  and  on  compounding  these  according  to  the  rules 
already  explained,  the  Ime  of  action  of  the  resviltant  determines 
the  line  along  which  the  total  weight  will  act.  Thus  the  center 
of  gravity  of  the  whole  system  of  bodies  will  be  the  centroid  of 
the  centers  of  gravity  of  the  separate  bodies,  weighted 
according  to  the  masses  of  these  bodies. 

90.  For  instance,  let  us  suppose  we  require  to  find  the 
center  of  gravity  of  a  pendulum  which  consists  of  a  wdre 
of  length  I  and  weight  w;  to  which  is  affixed  a  circular  bob 
of  weight  W,  the  center  of  the  circle  being  at  a  distance  a 
from  the  end  of  the  wire.  Let  AB  be  the  wire,  C  the  center 
of  the  bob,  and  D  the  middle  point  of  the  wire.  The  center 
of  gravity  of  the  wire  will  be  at  D  and  that  of  the  bob  at  C, 
so  that  the  center  of  oravitv  of  the  whole  will  be  at  the  centroiil 


O 

B 
Fig.  65 


120  CENTER  OF   GRAVITY 

of  the  points  D  and  C,  these  being  weiglited  in  the  ratio  w.W 
Denoting  this  center  of  gravity  by  G,  we  have,  from  the  formula 


X  = 


on  treating  the  line  ADCB  as  tlie  axis  of  x  and  taking  A  as  origin, 
.^       WAC+w-AD 

ALr  =  

W  +  w 

_  W{1  -a)  +  \  id 

EXAMPLES 

1.  Weights  of  3  pounds  are  placed  at  each  of  three  corners  of  a  square,  and 
a  weight  of  5  pounds  at  the  fourth  corner.    Find  their  center  of  gravity. 

2.  From  one  corner  of  a  cardboard  square  of  edge  6  inches,  a  square  of  edge 
3  inches  is  cut.    Find  the  center  of  gravity  of  the  remainder. 

3.  A  thin  rod  of  weight  6  ounces  and  length  6  inches  is  nailed  on  to  a  circle 
of  cardboard  of  weight  6  ounces  and  radius  6  inches  so  that  its  two  ends  are  on 
the  circumference  of  the  circle.    Find  the  center  of  gravity  of  the  whole. 

4.  A  bicycle  wheel  of  diameter  26  inches  weighs  3  pounds.  Each  spoke  is  of 
length  11  inches,  and  starts  from  the  hub  at  a  distance  of  \  inch  from  the  cen- 
tral axis  of  the  wheel.  If  one  spoke  is  taken  out,  find  the  center  of  gravity  of 
the  wheel. 

5.  A  hammer  has  for  handle  a  wooden  cylinder,  length  8  inches,  radius 
I  inch,  weight  8  ounces,  and  for  head  an  iron  cylinder,  from  which  a  hollow  is 
cut  into  which  the  handle  exactly  fits,  the  handle  coming  through  so  as  to  be 
exactly  fiu.sh  with  the  iron.  The  head  is  of  length  3  inches,  radius  1]  inch,  and 
weight  3  pounds.    Find  the  approximate  position  of  the  center  of  gravity. 

6.  A  box,  without  lid,  is  made  of  1-inch  wood  so  as  to  have  internal  dimen- 
sions 12  X  12  X  12  inches.    Find  the  position  of  its  „center  of  gravity. 

7.  A  uniform  thin  rod  28  inches  in  length  is  bent  so  that  the  two  parts,  of 
lengths  12  and  10  inches,  are  at  right  angles  to  one  another.  Find  the  center  of 
gravity. 

8.  A  uniform  wire  is  bent  into  the  form  of  a  triangle.  Show  that  the  center 
of  gravity  of  the  wire  coincides  with  the  center  of  the  circle  inscribed  in  the 
triangle  formed  by  joining  the  middle  points  of  the  sides. 

9.  A  T-square  is  made  of  cedar  of  uniform  density,  the  crosspiece  being  of 
dimensions  6  x  2  x  |  inches,  and  the  arm  being  of  dimensions  8  x  H  X  |^  inches. 
The  crosspiece  is  cut  away  so  that  the  under  surface  of  the  instrument  is 
plane.    Find  the  position  of  the  center  of  gravity  of  the  whole. 


TRIANGLE  121 

10.  Three  beads  of  weights  Wa,  Wb,  W^  are  placed  on  a  circular  wire,  and 
when  the  beads  are  at  the  points  A,  B,  C  on  the  circle,  the  center  of  gravity 
of  the  whole  is  found  to  coincide  with  O,  the  center  of  the  circle.    Show  that 

Wa  W,  W, 


sin  BOC      sin  COA      sin  AOB 

Center  of  Gravity  of  a  Lamina 

91.  It  is  often  of  importance  to  be  able  to  find  the  position 
of  the  center  of  gravity  of  a  lamina,  i.e.  of  a  thin,  plane  shell  of 
vmiform  thickness  and  density,  such,  for  instance,  as  is  obtained 
by  cutting  a  figure  out  of  a  sheet  of  cardboard. 

92.  Center  of  gravity  of  a  triangle.  Let  ABC  represent  a 
triangular  lamina  of  which  it  is  required  to  find  the  position  of 
the  center  of  gravity.  Let  us  imagine  the  triangle  divided  by 
lines  parallel  to  the  base  BC  into  a  very 
great  number  of  infinitely  narrow  strips. 
Let  2^(1  l^e  any  single  strip.  Since,  by 
hypothesis,  we  may  regard  this  strip  as 
of  vanishingly  small  width  and  thick- 
ness, w^e  may  treat  it  as  a  thin  rmiform 
rod.  The  center  of  gravity  of  a  thin  B 
uniform  rod  is  at  its  middle  point,  so 
that  the  weight  of  the  strip  pj  may  be  supposed  to  act  at  r,  the 
middle  point  of  2J(l- 

The  weights  of  the  other  strips  may  be  treated  in  the  same 
way,  so  that  the  weight  of  the  whole  triangle  may  be  replaced  by 
the  weights  of  a  system  of  particles  situated  at  the  middle  points 
of  these  strips. 

Now  if  D  is  the  middle  point  of  the  base  BC,  the  middle  points 
of  all  the  strips  lie  in  the  line  AD.  Thus  the  weight  of  the  tri- 
angle is  replaced  by  the  weights  of  a  number  of  particles,  all  of 
which  are  situated  in  the  line  AD.  It  follows  that  the  center  of 
gravity  of  the  whole  triangle  nnist  lie  in  the  line  AD. 

We  might  equally  well  have  supposed  the  triangle  divided  into 
strips  parallel  to  the  side  AC.    We  should  then  have  found  that 


122  CENTEK  OF  GEAYITY 

the  center  of  gravity  must  lie  in  the  line  BE,  where  E  is  the 
middle  point  oi  AC. 

These  two  results  fully  fix  the  position  of  tlie  center  of  gravity ; 
it  must  be  at  the  intersection  of  the  lines  AD,  BE. 

Join  DE.  Then  the  triangles  DCE,  BCA  are  two  similar 
triangles,  the  former  being  just  half  the  size  of  the  latter.  Thus 
DE  must  be  parallel  to  AB,  and  of  half  the  length  of  AB. 

It  now  follows  that  DGE,  AGB  are  similar  triangles,  of  which 
the  former  is  half  the  size  of  the  latter. 
Hence  G^D  is  half  of  ^G^. 

Thus  G  divides  AD  m  the  ratio  2:1. 
If  we  join  C  to  F,  the  middle  point  of 
AB,  we  can  in  the  same  way  show  that 
CF  must  divide  AD  in  the  ratio  2  : 1. 
Thus  CF  must  also  pass  through  G. 

The  three  lines  AD,  BE,  CF,  wliich 
join  the  vertices  of  the  triangle  to  the 
middle  points  of  the  opposite  sides,  are  called  the  medians  of  the 
triangle.  We  have  shown  that  the  three  medians  meet  in  the  same 
point  G,  and  that  this  point  is  the  center  of  gra\dty  of  the  triangle. 
We  have  also  shown  that  the  center  of  gravity  divides  any  median 
in  the  ratio  2  : 1,  i.e.  that  it  is  one  third  of  the  w^ay  up  the  median, 
starting  from  the  l)ase. 

93.  Center  of  gravity  of  any  polygon.  The  center  of  gravity 
of  any  rectilinear  polygon  can  be  found  by  dividing  it  up  into 
triangles  and  replacing  each  triangle  by  a  particle  at  its  center  of 
gravity. 

EXAMPLES 

1.  Show  that  the  center  of  gravity  of  a  triangle  coincides  with  that  of  three 
equal  particles  placed  at  its  angular  points. 

2.  Prove  that  if  the  center  of  gravity  of  a  triangle  coincides  with  its  ortho- 
center  the  triangle  is  equilateral. 

3.  A  cardboard  square  is  bent  along  a  diagonal  until  the  two  parts  are  at 
right  angles.     Find  the  position  of  its  center  of  gravity. 

4.  A  quarter  of  a  triangular  lamina  is  cut  off  by  a  line  parallel  to  its  base. 
Wliere  is  the  center  of  gravity  of  the  remainder? 


EOD   OF  VAKVING   DENSITY  123 

5.  A  right-angled  isosceles  triangle  is  cut  out  from  a  lamina  in  the  shape  of 
an  equilateral  triangle  so  as  to  have  the  same  base  as  the  original  triangle.  Find 
the  center  of  gravity  of  the  V-shaped  piece  left  over. 

6.  The  center  of  gravity  of  a  (juadrilatei-al  lies  on  one  of  its  diagonals.  Show 
that  this  diagonal  bisects  the  other  diagonal. 


Centers  of  Gravity  obtained  by  Integration 

94.  Center  of  gravity  of  a  rod  of  varying  density.  Let  AB  be 
a  rod  of  which,  the  weight  per  unit  length  varies  from  point  to 
point,  and  let  p  denote  the  weight  per  unit  length  at  any  point. 

Let  F,  Q  be  two  adjacent  points,  the  distances  of  P,  Q  from  the 
point  A  being  x  and  x  -{-  dx  respectively.  Then  the  length  PQ  is 
dx,  and  its  mass  is  pdx,  where  p  is  the    .  po  n 

mass  per  unit  length  at  this  point.  •"• 

When  dx  is  made  vanishingly  small,  ^^^-  ^^'^ 

the  distance  of  the  center  of  gravity  oi  PQ  from  A  may  be  taken 
to  be  X.  Hence  if  x  denotes  the  distance  of  the  center  of  gravity 
of  the  whole  rod  from  A, 

^iiiix) 


X< 


where  m  is  the  mass  of  any  element  such  as  PQ,  and  the  summa- 
tion is  taken  over  all  the  particles  of  which  the  rod  is  formed. 
Putting  m  =  p  dx,  this  becomes 

y\{pxdx) 


^{pd,S) 
or,  m  the  notation  of  the  integral  calculus, 

px  dx 


/' 


[pdx 


(28) 


■where  the  integration  extends  in  each  case  over  the  whole  rod. 
The  variable  p  will  be  a  function  of  x,  and  the  integrations  cannot 
be  performed  until  the  exact  form  of  this  function  is  known. 


124  CENTER   OE   GRAVITY 

95.  To  take  a  definite  instance,  let  us  suppose  that  the  density 
increases  uniformly  from  one  end  to  the  other.  Let  the  density  at 
A  be  0,  and  that  at  B  be  k.    If  the  rod  is  of  length  a,  the  density 

at  a  distance  x  from  A  will  be  k  |  -  V    Thus  we  must  put 
in  formula  (28),  and  so  obtain 


/'{= 


'  dx 


where  the  integration  is  from  a;  =  0  to  ic  =  a.    Dividing  numerator 
and  denominator  by  ->  we  obtain 


_   I 


JU    (a/JL> 


f 


t/y  CLiAj 


_^_2 
~'^a'      3  "*' 

showing  that  the  center  of  gravity  is  two  thirds  of  the  way  along 
the  rod. 

96.  We  can  use  this  result  to  obtain  the  center  of  gravity  of  a 
triangle.  As  in  §  92,  we  divide  the  triangle  into  parallel  strips, 
and  replace  each  strip  by  a  particle  at  its  middle  point.  The  mass 
of  each  particle  must  be  that  of  the  strip  which  it  replaces,  and 
this  is  jointly  proportional  to  the  width  and  the  length  of  the 
strip.  If  X  is  the  distance  of  any  particle  from  A  measured  along 
the  median  A  I),  the  width  of  a  strip  is  proportional  to  dx,  the 
length  intercepted  on  the  median,  while  the  length  of  the  strip  is 
])roportional  to  x,  the  distance  from  a.    Thus  p  dx  must  be  simply 


CIRCULAR  ARC 


125 


proportional  to  xdx,  and  as  we  have  just  found,  this  at  once  leads 
to  the  result  -_  _  ., 

where  a  is  the  length  of  the  median.  This  is  exactly  the  result 
previously  ohtamed. 

97.  Center  of  gravity  of  a  circular  arc.  The  same  method  can 
be  used  to  find  the  center  of  gravity  of  a  wire  bent  into  the  form 
of  a  circular  arc  PQ.  Let  0  be  the  center  of  the  circle,  and  A  the 
middle  point  of  the  arc,  and  let  the  whole  arc  subtend  an  angle 
2  a  at  the  center.  Consider  a  small  element  cd  of  the  half  PA  of 
the  wire.  Let  the  angle  dOA  1  je  6, 
and  cOA  be  ^  +  dd,  so  that  the 
element  subtends  the  angle  dd  at 
the  center.  If  a  is  the  radius  of 
the  circle,  the  length  of  this  ele- 
ment is  a  dd,  so  that  if  lo  is  the 
mass  of  the  wire  per  unit  length, 
the  mass  of  the  element  will  be 
wa  dd.  This  and  the  similar  ele- 
ment c'd'  in  the  half  AQ  oi  the 
wire  form  a  pair  of  equal  particles 
equidistant  from  the  central  line 
OA.    They  may  be  replaced  by  a 

single  particle  of  mass  2  wa  dd  at  their  center  of  gravity.  This 
center  of  gravity  is  in  OA,  at  the  point  at  which  the  line  joining 
the  two  elements  meets  OA,  and  hence  at  a  distance  a  cos  d  from  0. 
Denoting  this  by  x,  and  the  mass  2  wa  dd  by  m,  we  have,  for  the 
distance  x  of  the  center  of  gravity  of  the  whole  wire  from  0, 

'^{mx)  ^, 

X  =  : 


Fig.  69 


fa  COS d{2wadd) 
Clwadd 


126  CEIsTER  OF   GKAVITY 

where  the  integration  is  from  ^  =  0  to  6  —  a.    Simplifying,  we  find 


a  I         cos  Odd 

)  =  0 


r 


dd 


a  sma 


(29) 


giving  the  position  of  the  center  of  gravity. 

When  a  is  very  small,  sin  a  and  a  become  equal,  so  that  for 
very  small  values  of  a,  formula  (29)  reduces  to  x  =  a,  as  it  ought. 
This  simply  expresses  that  as  the  curvature  of  the  arc  decreases, 
the  center  of  gravity  approximates  more  and  more  closely  to  the 
middle  point  of  the  arc.  Finally,  when  a  =  0,  the  arc  becomes  a 
straight  rod,  and  the  center  of  gravity  is,  of  course,  found  to  be 
exactly  at  the  middle  point. 

For  an  arc  bent  into  a  semicircle,  we  take  a  =  —>  and  obtain 

a  sm  — 

2       2a 
X  = =  —  =  .6366  a. 

.      TT  TT 


98.  The  center  of  gravity  of  a  cir- 
cular arc  PQ  can  be  found  in  an  in- 
teresting manner,  without  the  use  of 
the  integral  calculus. 

From  symmetry  it  is  clear  that  the 
center  of  gravity  of  the  arc  AP  must 
lie  in  the  radius  which  bisects  the 
angle  A  OP.  Let  p  be  this  center  of 
gravity,  and  let  q  be  the  center  of 
gravity  of  the  arc  AQ.  Then  the 
center  of  gravity  or  the  whole  arc  P(2 
must  be  N',  the  middle  point  of  pq. 

Now  since  the  angle  pON  =  I  a, 
we  have 

ON  =  Op  cos  J  a . 


Fig.  70 


CIRCULAR  ARC  127 

This  relation  shows  tliat 

(the  distance  of  c.  g.  of  arc  2  a  from  center) 

=  cos  —  X  (the  distance  of  c.  g.  of  arc  a  from  center). 
Similarly 

(the  distance  of  c.  g.  of  arc  a  from  center) 

=  cos  —  X  (the  distance  of  c.  g.  of  arc  —  from  center), 
and  so  on.     Continuing  in  this  "vvay,  and  substituting,  we  obtain 

(the  distance  of  c.  g.  of  arc  2  a  from  center) 

a          a          a                 a 
=  cos  —  ■ cos  —  • cos  —  . • . cos 

2  -i  8  2"  +  ^ 

X  (the  distance  of  c.  g.  of  arc  —  from  center). 

If  we  make  n  very  great,  the  value  of  —  becomes  zero.     Thus  the  dis- 

a  " 

tance  of  the  c.  g.  of  an  arc  —  from  the  center  becomes  equal  to  a,  the  radius 

of  the  circle.     jNIaking  ?;  infinite,  we  have 

(the  distance  of  c.  g.  of  arc  2  a  from  center) 

=  a  cos  —  cos  —  cos  —  .  •  •  to  infinity. 
2         4         8  -^ 

T,T                                         a        sin  a 
^ow  cos  —  = , 


o 


a 


a    ■     a 

2  sin  — 

o 

cos-  = -'  etc., 

4        1    •     a 

^       4  sm  — 

4 


a 


.1         ,  C(  LL  CC 

so  that  cos  —  •  cos  —  •  cos  — 

2  4  8 


Makins:  n  infinite,  the  value  of  sin—  becomes  identical  with  -— ,  so 
a  '  2"  2 

that  2"  sin  —  becomes  identical  with  a,  and  we  have 

a         a         a        j.     •   i-    -t.         ^i'^  ^ 

cos  —  cos  —  cos  —  • . .  to  mtinitv  = > 

2         4         8  -^  a 

so  that  the  distance  of  the  c.  g.  of  the  arc  2  a  from  the  center  is  found  to 

,         sin  a  ,     „ 

be  a ,  as  before. 


128 


CE^^TER  OF  GRAVITY 


99.  Center  of  gravity  of  a  segment  of  a  circle.    Suppose  next 
that  we  require  to  find  the  center  of  gravity  of  a  segment  PAQN 

of  a  circle,  cut  off  by  a  chord  PNQ, 
which  subtends  an  angle  2  a  at 
the  center  0  of  the  circle.  Let  us 
divide  the  whole  segment  into  thin 
strips  parallel  to  the  chord,  and 
let  cc'dd'  in  fig.  71  be  a  typical 
strip  bounded  by  chords  cc',  dd'. 
Let  the  angle  cOA  be  6,  and  let 
do  A  be  d  +  dd.  Then  the  width 
of  the  strip  is  cd  sin  6  or  a  sin  6  dd, 
while  its  length  is  2  en  or  2  a  sin  ^. 
Thus  the  area  is  2  a  sin^  6  dd.  Its 
mass  may  be  supposed  to  be  all 
concentrated  at  n,  of  which  the 
distance  from  0  is  a  cos  6. 
Thus  if  X  is  the  distance  from  0  of  the  center  of  gravity  of  the 
whole  segment,  we  shall  have 

C{aco^e){2a^m''dde) 

x  = , 

/  {2  a  ^m^  Odd) 

and  the  integration  has  to  l)e  taken  from  ^  =  0  to  ^  =  a.    Simpli- 
fying, we  have 

I     sin"  6  cos  6  dd 

_  i/   0 


Fig.  71 


J  0 


sin-dde 


1  oi'n^  1 


sm  a 


i  (a;  —  sin  a  cos  a) 

2  sin^  a 

=  —a 


3     a  —  sin  a  cos  a 


SECTOR  OF  A  CIRCLE  129 

We  find,  on  i)uttin<>'  a:  =  —  >  that  the  center  of  <fravit.v  of  a  semi- 

4 
circle  is  at  a  distance  - —  a  froin  the  center. 

3  TT 

100.  Center  of  gravity  of  a  sector  of  a  circle.  The  center  of 
gravity  of  a  sector  of  a  circle  can  be  found  by  regarding  the  sector 
as  made  up  of  a  triangle  and  a  segment.  The  center  of  gravity  of 
the  triangle  and  of  the  seg- 
ment both  being  known,  it 
is  easy  to  find  the  center  of 
gravity  of  the  whole  figure. 

A    simpler    way    is    the 
following :    "VVe    can   divide 
the    sector  by   a    series    of 
radii  into  a  great   number  o* 
of    very    narrow    triangles. 
The  weight  of  each  triangle 
may    be    replaced    by    the 
weight  of  a  particle  placed 
at    its    center    of    gravity. 
Now,  in  the  limit,  when  the 
triangles  become  of  infini- 
tesimal width,  the  center  of  ^^'  *- 
gravity  of  each  is  on  its  median  at  a  distance  from  the  center  of 
the  circle  equal  to  ^  a,  where  a  is  the  radius  of  the  circle.     Thus 
all  the  particles  lie  on  a  circle  of  radius  |  a. 

The  weight  of  any  particle  must  be  equal  to  the  weight  of  the 
triangle  OPQ  which  it  replaces.  It  must,  therefore,  be  propor- 
tional to  the  base  PQ  of  the  triangle,  and  this  again  is  in-oporiional 
^)___-^P  to  j'j^Z'  ^1^6  piece  of  the  circle  of 
radius  1^  a  which  is  inclosed  by 
the  triangle.  Thus  the  weight  of 
the  particle  which  has  to  be 
placed  in  the  small  element  j^Q  f'l  this  circle  is  pro]iortional  to  the 
length  pq.  On  passing  to  the  limit,  and  making  the  number  of 
triangles   infinite,  we    find   that   the   string   of   particles   may  be 


lao 


CENTER   UE   GRAVITY 


replaced  by  a  wire  of  unifurm  density.  The  center  of  gravity  of 
this  wire  has  ah'eady  been  determined.  If  2  a  is  the  angle  of  the 
wire,  the  center  of  gravity  lies  on  the  radius  to  the  middle  point 

2  sin  a  ..         ^, 

of  the  wire  at  a  distance  -a irom  the  center. 

3  a 

Thus  the  center  of  gravity  of  the  original  sector  of  a  circle  of 
radius  a  and  angle  2  a  is  found  to  lie  on  the  central  radius  of  the 

sector  at  a  distance  -  a from  the  center. 

3        a 

101.  Center  of  gravity  of  a  spherical  cap.    The  piece  cut  off 

from  a  spherical  shell  by  a  plane  is  called  a  spherical  cap. 

The  center  of  gravity  of  a  spherical  cap 
cut  from  a  uniform  shell  can  easily  be 
found  by  the  methods  already  explained. 
Let  FQ  be  the  spherical  cap,  0  being 
the  center  of  the  sphere  from  which  it 
is  cut.  Let  0^  be  the  radius  perpen- 
dicular to  the  plane  FQ  by  which  the 
cap  is  bounded,  and  let  a  denote  the 
radius  of  the  sphere. 
Any  plane  parallel  to  FQ  will  cut  the  sphere  in  a  circle  of 

which  the  center  will  lie  on  OU.  Hence  by  taking  a  great  num- 
ber of  planes  parallel  to  FQ,  we   can 

divide  the  spherical  cap  into  a  number 

of  narrow  circular  rings,  each  having  its 

center  on  the  line  OU.    Let  us  consider  a 

single  circular  ring  cut  off  by  the  planes 

J  a  A',  BhB'.    Let  the  angles  AOE,  BOE  q 

be  equal  to  6  and  6  -\-  dd  respectively,  so 

that  the  ring  itself  subtends  an  angle  d6 

at  the  center.    The  width  AB  of  the  ring  is 

a  dd.    Its  circumference  may,  in  the  limit, 

be  supposed  equal  to  the  circumference 

of  the  circle  AaA'.     Since  Aa  =  a  sin  6, 


Fig. 


this  circumference  is  2  ira  sin  6. 


B' 

Fig.  75 

Thus  the  rins  under  consideration 


SPHEKICAL   CAP  A:NI>   JJELT 


131 


may  be  regarded  as  a  narrow  strip  of  length   2  vr  a  sin  d  and  of 
width  a  dO.    Its  area  is  accordingly  2  ira^  sin  6  cW. 

When  cW  is  made  very  small,  the  arc  BA  may  l)e  regarded  as  a 


e\\[i\\()E.    Thus 
-6 


straight  line  of  length  a  dd,  making  an  angle  — 

the  length  of  ha,  the  projection  of  BA  on  OE,  is  add  con 
or  a  sin  6  dd.    The  area  of  the  ring  BA  is  now  seen  to  be 

=  2  ira'  sin  6  dd 
=  2  7ra  •  ha. 

Thus  the  mass  of  the  ring  is  the  same  as  the  mass  of  the  ele- 
ment ha  of  a  rod  OE,  if  tliis  rod  is  of  uniform  density  such  that 
its  mass  per  unit  length  is  that  of  an  area  2  ira  of  the  shell.  The 
center  of  gravity  of  the  ring  we  have  Ijeen  considering  clearly  lies 
on  the  axis  OE,  so  that  in  finding  the  center  of  gravity  of  the 
spherical  cap  this  ring  may  obviously  be  replaced  by  the  element 
ha  of  this  rod. 

In  the  same  way  each  small  ring  may  be  replaced  by  the  cor- 
responding element  of  the  rod.  Thus  the  whole  cap  may  be 
replaced  by  the  length  rE  of  the  rod  (fig.  74)  wliich  is  inter- 
cepted between  the  boundary-plane  FQ  and  the  sphere.  Smce 
the  rod  is  uniform,  the  center  of  gravity  of  the  portion  rE  of 
the  rod  is  at  its  middle  point.  This  point  is  therefore  the  center 
of  gravity  of  the  spherical  cap. 

102.  Center  of  gravity  of  a  belt  cut  from  a 
spherical  shell  by  two  parallel  planes.  In  the 
same  way  we  can  find  the  center  of  gravity  of 
the  belt  cut  off  from  a  uniform  spherical  shell 
by  two  parallel  planes.  In  fig.  76  let  FQ,  F'Q' 
be  the  two  planes.  Then  we  can  divide  the 
belt  into  narrow  rings  by  planes  parallel  to 
FQ.  Each  ring,  as  before,  may  be  re])laced  by 
the  corresponding  element  of  a  uniform  rod 
along  the  axis  OE,  so  that  the  whole  belt  may 
be  replaced  by  the  portion  rr'  of  this  rod,  the 
portion  intercepted  between  the  two  planes  P(), 
F' Q'.    The  center  of  gravity  is  now  seen  to  be  the  middle  ]ioint  of  /•/. 


Fio.  7(> 


132 


CENTER  OF  GRAVITY 


Center  of  Gravity  of  a  Solid 

103.  Center  of  gravity  of  a  pyramid  on  a  plane  base.    Let  a 

pyramid  be  formed  having  any  plane  figure  OFQR  as  base  and 

anv  point  A  as  vertex.    We  can  find  the  center  of  gravity  of  a 

homogeneous  pyramid  by  dividing  it  into  thin  layers  parallel  to 

its  base,  by  a  series  of  parallel  planes. 

Let  opqr  be  any  such  layer,  this  layer  being  regarded  as  an 

infinitely  tliin  lamina.  Let  G  be  the  center  of  gravity  of  a  uni- 
form lamina  coinciding  with 
tlie  base  OPQR,  and  let  the 
line  AG  meet  the  lamina  opqr 
in  g.  Then,  from  tlie  geometry 
of  similar  figures,  it  is  clear 
that  g  occupies  a  position  in 
the  lamina  opqr  wliich  corre- 
sponds exactly  with  that  occu- 
^^i  pied  by  the  point  G  in  the 
larnma  OPQR.  Thus  g  will  be 
the  center  of  gravity  of  the 
lamina  opqr.  The  mass  of  this 
lamina    may,    accordingly,    be 

replaced  by  the  mass  of  a  single  particle  at  g. 

In  the.  same  way  each  of  the  laminas  mto  which  we  are  sup- 
posing  the   pyramid   to   be   divided   may  be 

replaced   by   a   smgle    particle    at    the   point 

at  which  the  lamina  intersects  the  line  AG. 

Thus  the  whole  pyramid   may  be  supposed 

replaced  by  a  series  of  particles  lying  along 

AG.    These  form  a  rod  of  varying  density, 

and  the   center  of   gravity  of   the  pyramid 

will  coincide  with  that  of  this  rod. 

The  center  of  gravity  of  the  rod  may  be 

found   by  the   method   already  explained  in 

§  94.     Consider    the    lamina   which    lies    between    two    adjacent 


Fig.  77 


Fig.  78 


PYRAMID  133 

parallel  planes  meeting  AG  in  fj,  (/'  respectively.  Let  Ag  =  x  and 
Ag'  =  X  +  dx,  so  that  the  lamina  intercepts  a  length  dx  on  A  G. 

Let  6  be  the  angle  between  A  G  and  the  per[)endicular  from  A 
on  to  the  base  of  the  lamina.    Then  the  thickness  of  the  lamina 

=  gg'  cos  6  —  dx  cos  6. 

If  S  is  the  area  of  the  base  of  the  pyramid,  the  area  of  the 
lamina  under  discussion  is 

AG' 

for  the  areas  of  the  different  laminas  are  proportional  to  the  squares 
of  their  linear  dimensions.  Thus  the  volume  of  the  lamina  we  are 
considering  ,2 

=  — — -  S  dx  cos  6. 
AG"" 

If  this  is  to  be  replaced  by  a  particle  occupying  the  length  dx 
of  the  rod  AG,  the  density  of  the  rod  must  be 

„  s  cos  e 

p  =  x- 

^  AG' 

Thus  the  rod  AG  must  be  of  a  density  which  varies  as  the 
square  of  the  distance  (x)  from  the  end  (A). 

The  distance  of  the  center  of  gravity  of  this  rod  from  A  is  now, 
by  the  formula  of  §  94, 

I     px  dx 


J, 
I 

L 


p  dx 

A 


x^  dx 

A 

G 


\AG' 
^AG^ 

^AG. 


1^4 


CENTER  OF  GRAVITY 


Thus  the  center  of  gravity  of  the  pyramid  is  in  the  line  AG, 
three  quarters  of  the  way  down  from  A. 

104.  Center  of  gravity  of  the  sector  of  a  sphere.  AVe  can  now 
find  the  center  of  gravity  of  tlie  sector  of  a  sphere,  —  tlie  volume 
cut  out  of  a  solid  sphere  by  a  right  circular  cone  having  its  ver- 
tex at  the  center  of  the  sphere.  To  do  this  we  divide  the  base 
FQ  of  the  sector  into  a  number  of  small  elements  of  area,  and 
then  divide  the  volume  of  the  sector  into  a  number  of  pyra- 
mids of  small  cross  section  by  taking  these  elements  of  area  as 
bases  and  joining  them  to  the  common  vertex  0.  These  pyra- 
mids are  all  of  the  same 
height,  so  that  their  masses 
are  proportional  to  their 
bases.  The  center  of  grav- 
ity of  each  pyramid  is  three 
quarters  of  the  distance 
jjj  down  from  0  to  its  base, 
and  is,  therefore,  at  a  dis- 
tance from  0  equal  to  three 
(juarters  of  tlie  radius  of 
the  sphere.  Thus,  if  we  con- 
struct a  second  sphere  hav- 
ing 0  as  its  center  and  of 
radius  equal  to  three  quar- 
ters of  the  radius  of  the  original  sphere,  the  center  of  gravity 
of  each  small  pyramid  will  lie  on  this  new  sphere.  Each  pyra- 
mid may  be  replaced  by  a  particle  at  its  center  of  gravity,  so 
that  the  whole  spherical  sector  may  be  replaced  by  a  series  of 
particles  lying  on  this  sphere  and  forming  the  spherical  cap  peq 
(fig-  79). 

The  mass  of  each  pyramid  is  proportional  to  the  base,  and  this 
again  is  proportional  to  the  part  of  the  splierical  shell  peq  which 
is  intercepted  by  the  pyramid.  Thus  the  spherical  shell  peq  which 
is  to  replace  the  original  volume  must  be  supposed  to  be  of  uni- 
form density. 


Fig.  79 


SECTOR  OF  A  SPHERE 


135 


The  sector  of  a  sphere  OPQ  has  now  been  replaced  by  the 
uniform  spherical  shell  2^1^  ^^'^^  the  center  of  gravity  of  this  shell 
is  known  to  be  G,  the  middle  point  of  re  in  fig.  79.  This  point  G 
is,  accordingly,  the  center  of  gravity  required. 

If  the  semivertical  angle  of  the  cone  by  which  the  sector  is 
bounded  is  a,  and  if  a  is  the  radius  of  the  sphere,  we  have 

Oe  =  ^  a,  Or  =  ^  a  cos  a, 
so  that  OG  =  I  a  (1  +  cos  a). 

In  particvdar,  if  a  =  —>  the  sector  becomes  a  hemisphere,  and 

OG  =  \a. 

Thus  the  center  of  gravity  of  a  hemisphere  is  three  eighths  of 
the  way  along  the  radius  which  is  perpendicular  to  its  base. 


Center  of  Gravity  of  Areas  and  Volumes  obtained  by 
Direct  Integration 

105.  Center  of  gravity  of  a  lamina.  To  find  the  center  of 
gravity  of  a  lamina  of  any  shape  by  integration,  we  take  any  con- 
venient set  of  axes  Ox,  Oy  in  the  plane  of  the  lamina,  and  imagine 
the  lamina  divided  into  small  elements 
by  two  series  of  lines,  one  parallel  to  the 
axis  Ox,  and  the  other  parallel  to  the 
axis  Oy. 

Consider  the  small  rectangular  element 
for  which  the  values  of  x  for  the  two 
edges  parallel  to  Oy  are  x  and  x  +  dx, 
and  the  values  of  y  for  the  two  other 
edges  are  y  and  y  +  dy.  The  area  of 
tliis  element  is  dxdy,  so  that  if  p  is  the  mass  of  the  lamina  per  unit 
area  at  this  point,  the  mass  of  the  element  will  be  p  dxdy.  ]\I ore- 
over,  when  dx,  dy  are  made  vanishingly  small  in  the  limit,  the 
mass  may  be  treated  as  a  particle.  Thus  the  whole  mass  of  the 
lamina  may  be  regarded  as  the  masses  of  a  number  of  particles. 


Fig.  .so 


13G  CENTER  OF   GRAVITY 

In  §  86  we  obtained  for  the  center  of  gravity  of  a  number  of 
particles  the  formulae 

In  the  present  instance  these  become 

jjpxdxdi/  iipydxdy 


I  j  p  dxdij  I  j  p  dxdy 


(30) 


the  sign  of  summation  being  replaced  by  an  integration  which  is 
to  extend  over  the  whole  area  of  the  lamina. 

If  the  lamina  is  uniform,  the  value  of  p  is  constant,  so  that 

/  /  px  dxdy  —  pi  j  1^  dxdy, 

and  so  on,  and  on  dividing  throughout  by  p,  the  formuliE  reduce  to 

j    xdxdy  ydxdy 


I  I dxdy  I  I  dxdy 


106.  Center  of  gravity  of  a  solid.  To  find  the  center  of  gravity 
of  a  solid  we  divide  it  into  small  solid  elements  by  three  systems 
of  planes  parallel  to  the  three  coordinate  planes.  The  volume  of 
any  small  element  is  then  dxdydz,  and  its  mass  is  p  dxdydz.  The 
formuhe  of  §  86  now  give  the  coordinates  of  the  center  of  gravity 
in  the  form 

I  I  I  px  dxdydz  i  i  i  P!/  ^^•^f^!/<^^ 

^  =  — 7^~rr '     2/  —  —        '  etc.  {61) 

j  j  j  P  dxdydz  f  i  i  P  dxdydz 


1^'TEGKATION  FOiLMLL.l^  137 

If  the  solid  is  homogeneous,  p  is  constant,  and  the  foimuku 
become 


\  \  \x dxdydz  /  /  / ^ dxdydz 

x  = >       y  = 

I  /  /  dxdydz  III  dxdydz 


etc. 


107.  Use  of  polar  coordinates.  Any  other  system  of  coordinates 
can,  of  course,  be  used  for  tinding  a  center  of  gravity  by  integra- 
tion. Tlie  only  coordmates  besides  Cartesians  which  are  of  much 
use  for  this  purpose  are  polar  coordinates. 

We  can  find  the  center  of  gravity  of  a  lamina  in  polar  coordinates 
by  supposing  the  Cartesian  coordinates  x,  y  connected  with  the 
polar  coordinates  ?%  6  by  the  usual  transformation 

x  =  r  cos  Q,     y  =  r  sin  6. 

Formuhe  (31)  then  become 


r  cos  6  — 


fCp  (/•  cos  d)  ()'  drdd)        fCpr  cos  d  drdd 


rdd 


r  sin  (7  = 


/  \p  (r  drdd)  (  I pr  drc 

ffp  (/■  sin  d)  (r  drdO)         ffpr"^  sm  d  drdd 
fCp(rdrde)  ffprdrdd 

in  which  r,  6  are  the  polar  coordinates  of  the  center  of  gravity. 
On  dividing  corresponding  sides  of  these  equations,  we  can  obtain 
an  equation  giving  the  0  coordinate  alone,  namely 


sin  6  drdd 
tan  6  = 

cos  0  drdd 


138  CENTER  OF   GRAVITY 

Similarly  we  can  find  the  center  of  gravity  of  a  solid  in  three- 
dimensional  polars  by  supposing  the  polar  coordinates  r,  6,  cf)  con- 
nected with  X,  y,  z  by  the  usual  transformation 

X  =  /•  sin  Q  cos  ^,     11  =  1''  sin  Q  sin  ^,     z  =  r  cos  6. 
Using  this' transformation,  the  first  of  formulae  (31)  becomes 

\  ]  \p  {i'  sin  6  cos  <}))  {r"^  sin  6  drddd(f>) 

f  sin  6  cos  (})  = — 

jjjp{7-^smedrddd(j>) 

j  i  j  pr^  sin^  6  cos  ^  drd6dj> 
fffpr'  sin  6'  f/rf/6'fZ</) 

while  similarly  we  have,  from  the  remaining  two  formulae, 

I  I  I  pr^  sin^  ^  sin  </>  drddd(f> 

r  sin  ^  sin  ^  = ^r^-; >  (33) 

Ipr'  sin  6  drd6d(f) 


(32) 


pr'^  sin  ^  cos  6  drddd<^ 

rcos^  =  —  ^ (34) 

III  /37-"  sin  6  drdddxfy 

108.  An  exactly  similar  method  will  lead  to  formulae  giving 
the  position  of  the  center  of  gravity  in  any  system  of  coordinates. 

The  methods  which  have  already  been  employed,  or  a  combina- 
tion of  them,  will  suffice  to  determine  any  center  of  gravity.  As 
illustrations  of  the  use  and  combination  of  these  methods,  we 
shall  find  the  center  of  gravity  of  the  same  solid  figure  in  three 
different  ways. 


ILLUSTRATIVE  EXAMPLE 


139 


ILLUSTRATIVE  EXAMPLE 

A  right  circular  cone  OPQ  is  scooped  out  of  a  solid  homogeneous  sphere,  the 
vertex  of  the  cone  O  being  on  the  surface  of  the  sphere,  and  its  axis  being  a  diameter 
of  the  sphere.    It  is  required  to  find  the  center  of  gravity  of  the  remainder. 

Method  I.  Polar  coordinates.  First  let  us  use  polar  coordinates,  taking  the 
vertex  0  of  the  cone  as  origin,  and  the  axis  of  the  cone  as  initial  line.  If  a  is 
the  semivertical  angle  of  the  cone,  the  equation  of  the  cone  is  ^  =  a.  If  a 
is  the  radius  of  the  sphere,  the  equation  of  the  sphere  is  r  =  2  a  cos  6.  Tlie 
center  of  gravity  must  from  symmetry  lie  on  the  axis  0  =  0,  so  that  0  =  0,  and 
equation  (34)  becomes 


fffpr'^  sin  0  cos  0  drded(t> 


pr^  sin  0  drd0d4> 


The  solid  is  supposed  to  be  homo- 
geneous, so  that  p  is  a  constant,  and 
may,  therefore,  be  taken  outside  the 
sign  of  integration  in  both  numerator 
and  denominator.  The  limits  of  inte- 
gration for  0  are  from  (p  =  0  to  0  =  2  tt, 
so  that  this  integration  may  be  per- 
formed in  each  case.    Doing  this,  and  dividing  out  by  2  irp,  we  are  left  with 


^=0 


//• 


3  sin  0  cos  0  drdd 


SI' 


r2  sin  0  drdd 


We  may  next  integrate  with  respect  to  r,  the  limits  being  r=  0  to  r  =  2a  cos^, 
and  obtain 

A-  (2  a  cos  0)*  sin  0  cos  0  d0 

r  = 


C\{2acose)^sin0dd 

fcoss  0  sin  0  d0 
I 


cos^^sin^d^ 


The  limits   of  integration  for  0  are  obviously   from    0  =  a  (the  cone)  to 


0  =  —  (the  tangent  plane  to  the  sphere). 


We  have 


f''cos^0sin0d0  =  —  i[cos6e]"=  Jcos^a, 
f' cos3 0  sin  0d0  =  —  \  [cos* 0]' =  ^  cos*  a. 


14U 


CENTEll   OF   GKAVITY 


Substituting  these  values,  we  find 

h  cos*5  a 
r  —  ']a  ■ 


=  a  cos^a. 


Tluis  the  center  of  gravity  is  on  the  axis  of  the  cone  at  a  distance  a  cos^  a 
from  the  vertex.. 

Method  II.   Cartesian  coordinates.   We  may  next  employ  Cartesian  coordi- 
nates, taking  0  as  origin  and  the  axis  of  the  cone  as  axis  of  x.    The  equation  of 

the  cone  is  now 

2/2  +  2;2  =  x2tan2a, 
while  that  of  the  sjDhere  is 
X-  +  y^  +  z2  —  2ax  =  0. 
From  §  100,  we  have 


Iff- 


X dxdydz 


Fig.  82 


I  \  \  dxdydz 

In  each  integral  we  may  inte- 
grate first  with  respect  to  y  and  z 
together.    We  have  to  evaluate 
the  same  integral  in  both  cases,  namely  I  (dydz,  the  limits  being  given  by 

?/2  -|-  z-  =  a;2  tan2  a 
and  ?/2  +  z2  _  2  ax  —  x-. 

The  problem  is  the  same  as  that  of  finding  the  area  of  a  circular  ring  of 
inner  and  outer  radii  x  tan  a  and  V'2  ax  —  xP-  respectively.  (This  ring  is,  of 
course,  the  intercept  of  the  solid  on  the  plane  parallel  to  the  yz  plane.)  The 
area  of  the  ring  is 

■K  (2  ax  —  a;2)  —  TT  [x?  tan2  or)  =  tt  (2  ax  —  a:2  sec2  a), 

and  on  substituting  this  value  for  |  (dydz,  the  formula  becomes 

TTX  (2  ax  —  x?  sec2  a)  dx 


/' 


j  -77(2  ax  —  x^  sec2  a)  dx 

The  limits  of  integration  are  now  from  x  =  0,  the  origin,  to  x  =  2acos2cr, 
the  value  of  x  on  the  plane  PQ.  Evaluating  the  integrals,  and  substituting 
these  limits,  we  obtain 

_  _  2  aTT  ^  (2  a  cos2  a)^  —  tt  sec2  a  ]  (2  a  cos2  a)* 
2  air  \  (2  a  cos2  a  )2  —  tt  sec2  a  i  (2  a  cos^  a)^ 
—  a  cos2  cr, 
giving  the  same  result  as  before.  • 


ILLUSTKATIVE  EXxiMPLE 


141 


Method  III.   Geometrical  Method.    The  center  of  gravity  can  also  be  found 
by  regarding  the  given  volume  as  the  sums 
and  differences  of  simpler  volumes  of  which 
the  center  of  gravity  is  already  known.  /  ^P 

The  volume  is  obtained  by  taking  the 
complete  sphere  OPsQ  and  subtracting  from 
it  the  cone  OPrQ  and  the  spherical  segment  o\ 
PrQs.  The  center  of  gravity  of  the  sphere 
and  cone  are  known,  —  that  of  the  segment 
PrQs  is  most  easily  found  by  regarding  it  as 
the  difference  between  the  sector  CPsQ  and 
the  cone  CPrQ.  Thus  we  regard  the  original 
figure  as  made  up  of  '  yiq   gS 

(sphere  OPsQ)  -  (cone  OPrQ)  -  (sector  CPsQ)  +  (cone  CPrQ).     ■ 

The  volumes  of  these,  and  the  distances  of  their  centers  of  gravity  from  O 
measured  along  OC,  are  as  follows  : 


Figure 


Volume 


Distance  of  C.G.  from  O 


+  sphere                   |  -rra^  a 

—  cone  OPrQ           —  J  (2  a  cos- nr)  (jra- sin^  2a)  ^{2  a  cos- a) 

-  sector  CPsQ         —  |  ira^  (1  -  cos  2a)  a  +  J  a  (1  +  cos  2a) 
+  cone  CPrQ           i  (a  cos  2a)  (vra^  sin2  2a)  a +1  a  cos  2a 

In  this  table  the  negative  sign  denotes  that  a  figure  is  to  be  removed,  so  tliat 
its  volume  must  be  reckoned  as  of  negative  sign. 

Denoting  the  distance  of  any  center  of  gravity  from  O  by  x,  anil  using 
the  formula 


of  §  86,  we  obtain  as  the  distance  of  the  center  of  gravity  of  the  whole  figure 
from  0 

_  _         4 Tra*  —  \(2a cos^ a)-(Trn- sin'- 2a)  —  § 7ra*(l  —  cos 2a)  { 1  +  i (1  +  cos 2a) } 
|7ra3  — i(2acos2a)(7ra-sin'-22a)  —  j7ra-'(l  —  cos2a)  +  J(acos2a)(7ra-sin-2a) 

+  l(a  cos2a)(7ra2  sin"-2a)a(l-l- §  acos2a) 
|7ra3— J(2acos2a)(7ra2sin2  2a)  — |7ra^(l  — cos2a)  +  |(acos2a:)(7ra-siu2  2a) ' 

which,  after  reduction,  gives 


X  =  a  cos-'  a. 


tlie  same  result  as  before. 


142  CENTEK  OF  GEAVITY 

GENERAL  EXAMPLES 

1.  A  plane  quadrilateral  ABCD  is  bisected  by  the  diagonal  AC,  and 
this  diagonal  is  divided  in  the  ratio  a  :  6  by  the  diagonal  BD.  Prove 
that  the  center  of  gravity  of  the  quadrilateral  lies  in  AC  and  divides  ib 
into  two  parts  in  the  ratio  2a  +  h  •.2h  +  a. 

2.  A  uniform  wire  is  bent  into  the  form  of  a  circular  arc  and  the  two 
bounding  radii,  and  the  center  of  gravity  of  the  whole  is  found  to  be 
at  the  center.  Show  that  the  angle  subtended  by  the  arc  at  the  center  is 
tan-i(-l). 

3.  The  three  feet  of  a  circular  table  are  vertically  below  the  rim  and 
form  an  equilateral  triangle.  Prove  that  a  weight  less  than  that  of  the 
complete  table  cannot  upset  it. 

4.  A  triangular  table  is  supported  by  three  legs  at  the  middle  points 
of  its  sides,  and  a  weight  W  is  placed  on  it  in  any  position.  It  is  found 
that  the  table  will  just  be  upset  if  a  weight  P  is  placed  at  one  angular 
corner.  The  corresponding  weights  needed  to  upset  it  at  the  other  corners 
are  Q,  R.  Prove  that  P  +  Q  +  R  is  independent  of  the  position  of  the 
weight  IF. 

5.  Weights  are  nailed  to  the  three  corners  of  a  triangular  lamina,  each 
proportional  to  the  length  of  the  opposite  side  of  the  triangle,  and  of  com- 
bined weight  equal  to  the  original  weight  of  the"  lamina.  Show  that  the 
center  of  gravity  of  the  triangle  is  at  the  center  of  the  nine-point  circle. 

6.  A  uniform  triangular  lamina  of  w^eight  W  and  sides  a,  b,  c  is  sus- 
pended from  a  fixed  point  by  strings  of  lengths  /j,  l^,  I3  attached  to  its 
angular  points.    Show  that  the  tensions  of  the  strings  are 

Wkl^,     WH^,     WHs, 

where  k  =  [3  (If  +  l.r  +  l-s)  -  a^  -  b^  -  c^yK 

7.  Explain  how  a  clock  hand  on  a  smooth  jiivot  can  be  made  to  show 
the  time  by  means  of  watchwork,  carrying  a  weight  r^und,  concealed  in 
the  clock  hand. 

8.  A  spindle-shaped  solid  of  uniform  material  is  bounded  by  two  right 
circular  cones  of  altitudes  6  and  2  inches  with  a  common  circular  base  of 
radius  1  inch.  It  is  suspended  by  a  string  attached  to  a  point  on  the  rim 
of  the  circular  base.  Find  the  inclination  of  the  axis  of  the  spindle  to  the 
vertical  when  it  is  hanging  freely. 

9.  A  pack  of  cards  is  laid  on  a  table,  and  each  projects  beyond  the  one 
below  it  in  the  direction  of  the  length  of  the  pack  to  such  a  distance  that 
each  card  is  on  the  point  of  tumbling,  independently  of  those  below  it. 
Prove  that  the  distances  between  the  extremities  of  successive  cards  will 
form  a  harmonic  progression. 


EXAMPLES  143 

10.  Prove  that  the  center  of  gravity  of  any  portion  PQ  of  a  uniform 
heavy  string  hanging  freely  is  vertically  above  the  intersection  of  the 
tangents  at  P,  Q. 

11.  A  hemispherical  shell  has  inner  and  outer  radii  a,  h.  Show  that 
the  distance  of  its  center  of  gravity  from  its  geometrical  center  is 

3(a  +  i)(rt2  +  ft2-) 
8     a2  +  ah  +  6^ 

12.  An  anchor  ring  is  cut  in  two  equal  parts  by  a  plane  through 
its  center  which  passes  through  its  axis.  Find  the  center  of  gravity  of 
either  half. 

13.  Prove  that  the  pull  exerted  by  a  man  in  a  tug  of  war  is  -  of  his 

h 
weight,  where  a  is  the  horizontal  projection  of  a  line  joining  his  heels  to 
his  center  of  gravity,  and  b  is  the  height  of  the  rope  above  the  ground. 

14.  Prove  that  a  horse  weighing  W  pounds  can  exert  a  horizontal  pull 
of  Wa/k  pounds  at  a  height  h  above  the  ground  by  advancing  his  center 
of  gravity  a  distance  a  in  front  of  its  position  when  he  is  standing  upright 
on  his  legs. 

15.  A  rod  of  varying  density  and  material  is  sujjported  by  a  man's  two 
forefingers,  across  which  it  rests  in  a  horizontal  position.  The  man  moves 
his  fingers  toward  one  another,  keeping  them  in  the  same  horizontal  plane, 
and  allowing  the  rod  to  slip  over  one  or  both  of  his  fingers.  Show  that 
when  his  fingers  touch,  the  center  of  gravity  of  the  rod  will  be  between 
the  points  of  contact  of  his  fingers  with  the  rod. 

16.  A  semicircular  disk  rests  in  a  vertical  plane  with  its  curved  edge  on 
a  rough  horizontal  and  an  equally  rough  vertical  plane,  the  coefficient  of 
friction  being  p..  Show  that  the  greatest  angle  that  the  bounding  diameter 
can  make  with  the  vertical  is 

sin~^ • 

1  +  M'^  -2 

17.  A  hemisphere  of  radius  a  and  weight  W  is  placed  with  its  curved 
surface  on  a  smooth  table,  and  a  string  of  length  1(1  <  a)  is  attached  to  a 
point  on  its  rim  and  to  a  point  on  the  table.  Prove  that  the  tension  of  the 
string  is  ^  a-l 


8       V2  al  -  P 

18.  A  triangular  lamina  of  weight  W  is  supported  by  three  vertical 
strings  attached  to  its  angular  points  so  that  tlie  plane  of  the  triangle  is 
horizontal ;  a  particle  of  weight  W  is  placed  at  the  orthocenter  of  the 
triangle.    Prove  that  the  tensions  of  the  strings  are  given  by 

l  +  3cot5cotC      l  +  3cotCcot^      l  +  3cot^cot5       2 


144  CEKTEll   UF  GRAVITY 

19.  Find  the  center  of  gravity  of  a  lamina  bounded  by  a  parabola  and 
a  line  perpendicular  to  its  axis. 

20.  Find  tlie  center  of  gravity  of  the  volume  cut  from  a  solid  parabo- 
loid by  a  plane  perpendicular  to  its  axis. 

21.  Find  the  center  of  gravity  of  the  ai-ea  inclosed  by  two  radii  of 
an  ellipse. 

22.  Find  the  center  of  gravity  of  the  volume  cut  off  from  a  solid  ellipsoid 
by  a  i)Iane  through  the  center. 

23.  Find  the  center  of  gravity  of  half  of  an  elliiisoidal  shell,  this  being 
bounded  by  two  similar  concentric  and  coaxial  ellipsoids,  and  a  plane 
through  the  center. 

24.  A  right  circular  cone  whose  base  is  of  radius  r  is  divided  into  two 
equal  parts  by  a  plane  through  the  axis.     Prove  that  the  distance  of  the 

center  of  gravity  of  either  half  from  the  axis  is  —  • 

25.  Find  the  center  of  gravity  of  a  lamina  bounded  by  the  semicubical 
parabola  x^  =  ay'^,  the  axis  of  x,  and  the  ordinate  x  =  a. 

26.  Find  the  center  of  gravity  of  a  single  loop  of  the  curve 

r  =  a  sin  3^. 

27.  Find  the  center  of  gravity  of  an  octant  of  a  si)here. 

28.  A  cylindrical  hole  of  radius  a  is  drilled  through  a  hemisphere  of 
radius  b  so  that  the  radius  perpendicular  to  the  base  of  the  hemisphere  is 
also  the  central  line  of  the  hole.     Find  the  center  of  gravity  of  the  figure. 

29.  Find  the  center  of  gravity  of  the  area .  inclosed  between  the  two 

''"■'^^^^  x"-  +  f  =  a^;  x^  +  i/  =  2  ah. 

30.  Find  the  center  of  gravity  of  a  lens  made  of  homogeneous  glass, 
having  spherical  surfaces  of  radii  r,  s,  and  of  which  the  thickness  is  t  at 
the  center  and  zero  at  the  edse. 


CHAPTER   Vll 

« 

WORK 

109.  Measurement  of  work.  There  are  various  kinds  of  work, 
but  in  mechanics  we  are  concerned  only  with  the  work  done  in 
moving  bodies  which  are  acted  on  by  forces.  Such  work  is 
described  as  mechanical  work.  We  say  that  mechanical  work  is 
done  whenever  a  body  is  moved  in  opposition  to  the  forces  acting 
on  it,  as,  for  instance,  m  raising  a  weight,  in  dragging  a  heavy 
body  over  a  rough  surface,  or  in  stretching  an  elastic  string.  In 
the  first  case  work  is  performed  against  the  force  of  gravity,  in 
the  second  case  against  the  frictional  force  exerted  on  the  moving 
body  by  the  rough  surface,  and  in  the  third  case  against  the 
tension  of  the  string. 

Obviously  in  estimating  the  amount  of  work  done,  two  factors 
have  to  be  taken  into  account,  namely  the  amount  of  the  force 
actiag  on  the  body  and  the  distance  througli  which  the  body  is 
moved  in  opposition  to  this  force.  The  amount  of  work  will  clearly 
be  directly  proportional  to  the  force,  —  in  raising  a  weight  of  200 
pounds  through  a  given  distance  we  do  twice  as  much  work  as 
in  raising  a  weight  of  100  pounds  through  the  same  distance.  It 
will  also  be  proportional  to  the  distance  moved, —  in  raising  a 
weight  through  two  feet  we  do  twice  as  much  work  as  in  raising 
the  same  weight  through  one  foot.  Thus  the  amount  of  work  done 
varies  as  the  product  of  the  force  and  the  distance. 

The  amount  of  work  done  in  raising  a  weight  of  one  pound 
through  a  height  of  one  foot  is  called  one  foot  pound. 

From  what  has  been  said,  it  is  clear  lliat  the  work  done  in 
raising  a  weight  of  w  pounds  through  a  height  of  h  feet  is 
wh  foot  pounds.  Also,  the  work  done  in  mo\'ing  a  body  a  distance 
of  s  feet  in  opposition  to  a  force  of  F  pounds  weight  is  Fs  foot 

145 


146  WORK 

pounds.  Thus  we  may  say  that  the  work  done  in  moving  a  body 
through  any  distance  against  a  uniform  force  is  the  product  of  the 
distance  and  the  force. 

Suppose,  for  instance,  that  it  is  found  that  the  force  reqviired  to  drag  a 
railway  train  along  a  level  track  is  equal  to  the  weight  of  10,000  pounds, 
then  the  work  done  in  hauling  this  train  a  distance  of  100  miles 

=  100  X  5280  X  10,000  foot  pounds. 

110.  Rate  of  performing  work.  Work  frequently  has  to  be  done 
within  a  given  time,  so  tliat  it  is  often  necessary  to  measure  the 
rate  at  which  work  is  being  done.  The  rate  of  doing  work  in 
which  33,000  foot  pounds  are  done  per  minute  is  called  one  horse 
jpoioer  (1  H.  P.). 

This  unit  was  introduced  by  Watt,  and  was  supposed  to  measure  the 
rate  of  working  of  an  ordinary  horse.  It  is  found,  however,  that  very  few 
horses  are  capable  of  working  continuously  at  one  horse  power  for  any 
length  of  time. 

As  an  example  of  the  calculation  of  horse  power,  let  us  find  the  horse 
power  required  of  an  engine  to  haul  a  train  at  30  miles  an  hour,  the  fric- 
tional  resistance  being  equal  to  the  weight  of  10,000  pounds.  A  velocity 
of  30  miles  an  hour  =  44:  feet  per  second,  so  that  the  work  done  per  second 
=  44  X  10,000  foot  pounds.  Since  one  horse  power  =  550  foot  pounds  per 
second,  we  see  that  the  horse  power  required 

44  X  10,000 


550 


=  800  horse  power. 


This  gives  the  horse  power  required  to  haul  the  train  at  a  steady  speed 
of  30  miles  per  hour.  We  shall  find  that  if  the  speed  is  not  constant  the 
horse  power  will  be  different,  part  of  the  work  being  used  up  in  producing 
the  acceleration  of  the  motion.  For  the  present,  however,  we  confine  our 
attention  to  motion  with  uniform  velocity. 

Absolute  Unit  of  Work 

HI.  We  have  already  seen  that  besides  the  practical  unit  of 
force,  which  is  the  weight  of  a  unit  mass,  there  is  also  a  second 
unit  of  force,  known  as  the  absolute  unit,  which  is  defined  as  being 
a  force  capable  of  producing  unit  acceleration  in  unit  mass.  As 
the  practical  unit  produces  acceleration  g  in  unit  mass,  where  g  is 


MEASUPvEMEXT  AND   UNITS  147 

the  acceleration  clue  to  gravity,  it  follows  that  the  practical  unit 
is  g  times  the  absolute  unit. 

In  practical  British  units,  the  unit  force  is  tlie  pound  weight. 
In  absolute  units,  the  corresponding  unit  is  known  as  the  poundal; 
it  is  the  force  which  will  produce  unit  acceleration  in  a  mass  of 
one  pound. 

The  practical  unit  of  work,  as  we  have  said,  is  the  work  done  in 
raising  a  mass  of  one  pound  through  one  foot,  i.e.  in  moving 
through  one  foot  the  point  of  application  of  one  pound  weight. 
There  is  also  an  absolute  unit  of  work,  namely  the  work  done  in 
moving  through  one  foot  the  point  of  application  of  one  poundal. 
This  unit  is  called  the  foot  poundal.  Smce  one  pound  weight  is 
equal  to  g  poundals,  we  obviously  have  the  relation 

1  foot  pound  =  g  foot  poundals. 


EXAMPLES 

1.  At  what  speed  can  a  horse  of  1  horse  power  draw  a  cart  weighing  1  ton, 
friction  being  supposed  to  cause  a  liorizontal  force  equal  to  one  fortieth  of  the 
weight  of  the  cart  ? 

2.  A  body  resisted  by  a  force  of  P  poundals  is  moved  against  this  resistance 
with  a  velocity  v.    What  horse  power  is  required  ? 

3.  At  what  rate  can  a  steam  roller  of  7  horse  power  and  weight  1  ton  roll  a 
path,  the  resistance  due  to  friction  being  equal  to  the  weight  of  the  roller  ? 

4.  A  snail  weighing  \  ounce  climbs  a  wall  6  feet  in  height  in  4  hours.  At 
what  horse  power  is  he  working  ? 

5.  A  load  of  bricks  weighing  5  tons  has  to  be  raised  to  the  top  of  a  house 
50  feet  in  height  by  10  laborers,  each  of  whom  works  at  an  average  rate  of 
y'2  horse  power.    How  long  ought  the  job  to  take  ? 

6.  The  piston  of  an  engine  has  an  area  of  a  square  feet  and  a  stroke  of 
I  feet,  and  the  engine  makes  p  revolutions  per  minute.  If  the  pressure  per  unit 
area  acting  on  the  piston  is  p  pounds  weight  per  square  foot,  prove  that  the 
hoi-se  power  at  which  the  engine  is  working  is 

plan 
33,000 ' 

7.  A  locomotive  has  a  circular  piston  of  diameter  17  inches,  and  stroke  26 
inches.  It  makes  250  revolutions  per  minute,  the  pressure  being  'I'lb  pounds 
weight  per  square  inch.    Find  its  horse  power. 


148  WORK 

8.  If  200  horse  power  is  required  to  drive  a  steamer  150  feet  long  at  a  speed 
of  9  knots,  prove  that  25,G00  horse  power  will  be  required  to  drive  a  similar 
steamer,  GOO  feet  long,  similarly  immersed,  at  18  knots,  assuming  that  the 
resistance  is  proportional  to  the  wetted  surface  and  the  square  of  the  velocity 
through  the  water.  Prove  also  that  the  cost  of  coal  per  ton  of  cargo  will  be 
the  same  in  the  two  steamers. 

9.  Fifty  hor.se  power  is  transmitted  from  one  .shaft  to  another  by  means  of 
a  belt  moving  over  two  wheels  on  the  shafts  with  a  linear  velocity  of  250  feet 
per  miiuite.    Find  the  difference  of  tensions  on  the  two  sides  of  the  belt. 

10.  A  locomotive  consumes  1^  pounds  of  coal  per  horse-power-hour.  How 
much  coal  is  required  to  haul  a  train  of  total  weight  1000  tons  over  50  miles  of 
level  road  on  which  the  resistance  to  friction  is  12  pounds  weight  per  ton  ? 

11.  A  liner  of  22,000  horse  power  makes  a  run  of  3300  miles  in  six  days. 
Find  the  resistance  to  the  shij^'s  motion. 

Work  done  against  a  Vaeiable  Force 

112.  If  a  body  is  moved  in  opposition  to  a  force  which  is  not 
of  constant  intensity  but  varies  from  point  to  pomt  on  the  path 
of  the  moving  body,  we  can  no  longer  vise  the  formula  Fs  for  the 
amount  of  work  performed. 

To  calculate  the  amount  of  work  done,  we  divide  up  the  whole 
range  over  which  motion  takes  place  into  an  infinite  number  of 
infinitesimally  small  ranges,  each  of  these  ranges  being  so  small 
that  the  force  opposing  the  motion  may  be  regarded  as  of  constant 
magnitude  durmg  the  motion  through  any  one  of  them. 

If  ds  is  any  small  range  at  a  distance  s  from  the  starting  point, 
and  if  F  is  the  intensity  of  the  force  opposing  the  motion  while 
the  body  moves  through  the  small  range  ds,  then  the  work  done  in 
moving  through  tliis  range  is  Fds.  The  total  work  done,  the  sum 
of  the  amounts  of  work  done  in  all  the  ranges,  is,  accordingly, 


JFds. 


Work  done  in  stretching  an  Elastic  String 

113.  As  an  example  of  the  use  of  this  formula,  let  us  find  the 
work  done  in  stretching  an  elastic  string.  Let  the  natural  length 
of  the  string  be  I,  and  let  X  denote  its  modulus  of  elasticity. 


WORK  OF   STRETCHING  A   STRING  149 

When  the  leugth   of  the   string   is  x,  its  tension   T,  by  the 
formula  of  §  39,  is  given  by 

In  stretching  the  string  through  a  further  distance  dx,  —  i.e.  from 
length  X  to  length  x  +  dx,  —  the  work  done 

=  Tdx 

=  --(,«  —  /)  dx. 

By  integration,  we  find  that  the  work  done  in  stretching  a  string 
from  length  a  to  length  b 


-f 


—  {x  —  l)dx 


=  ^^(h  +  a-2/){h-a). 

The  distance  stretched  is  h  —  a,  while  —  {h  +  a  —  2  I)  is  the 

tension  when  half  of  the  stretching  has  been  completed,  i.e.  when 
X  =  I  (a  +  h). 

Thus  we  have  found  that 

The  tvork  done  in  stretclnufj  an  clastic  string  from  an//  length  a, 
greater  thctn  the  natural  length  of  the  string,  to  a  length  h,  is  cqnal 
to  the  tension  at  lengtli  \{a  +  h)  multiplied  hi/  {h  —  a). 

Obviously,  if  the  tension  is  measured  in  })ounds  weight  and  the 
extension  {h  —  «)  in  feet,  the  product  will  give  the  amount  of  work 
measured  in  foot  pounds.  If  the  tension  is  measured  m  pouiulals. 
and  (b  —  a)  in  feet,  the  product  will  give  the  amount  of  work  in 
foot  poimdals. 


150 


WORK 


AVOEK    REPRESENTED    BY    AN    ArEA 

114.  Let  PQ  represent  the  path  described  by  a  movmg  body, 
and  let  us  draw  ordinates  at  each  point  in  FQ  to  represent,  on 
any  scale  we  please,  the  force  opposing  the  motion  of  the  body  at 

that  point.  Let  s,  r  be  two  adja- 
cent poin^ts,  and  let  ss\  rr'  be 
the  ordinates  at  these  points. 

Then  the  area  of  the  small 
strip  ss'rr'  may,  in  the  limit,  be 
supposed  equal  to  sr  multiplied 
by  ss'.  On  the  scale  on  which 
we  are  representmg  forces,  this 
product  will  represent  the  dis- 
tance sr  multiplied  by  the  force  opposing  the  motion  of  the  body 
from  s  to  r.  In  other  words,  the  small  area  ss'rr'  will  represent 
the  work  done  in  moving  the  body  from  s  to  r. 

By  addition  of  such  small  areas,  we  find  that  the  complete  area 
PP'QQ'  represents  the  work  done  in  moving  from  P  to  Q. 

115.  This  method  gives  a  simple  way  of  investigating  the  work  done  in 
stretching  an  elastic  string,  already  calculated  in  §  113.  Let  OP  be  the 
natui-al  length.  For  the  sake  of  definiteness  suppose  that  the  end  0  is 
held  fast,  and  that  as  the  string  is  stretched  the  point  P  moves  along  the 
line  OP.  Let  it  be  required  to  find  the  work  done  in  stretching  the  string 
from  a  length  OA  to  a  length  OB. 

Let  Q  be  any  point  of  the  line  Bi 

OPAB,  and  let  QQ'  be  drawn  to  ^  O' 

represent  the  tension  when  the  - 

length  of  the  string  is  OQ. 

For  different  positions  of  Q,  the 
ordinate  QQ'  will  be  of  different 
heights.  Since,  by  Hooke's  law, 
the  tension  is  proportional  to  the  extension,  the  height  of  the  ordinate 
QQ'  (representing  the  tension)  will  always  be  in  the  same  ratio  to  PQ  (the 
extension).  Thus  Q'  is  always  on  a  certain  straight  line  through  P.  If  A  A', 
BE'  are  tlie  ordinates  which  represent  the  tensions  at  .4 ,  B,  this  line  will, 
of  course,  pass  through  the  points  A',  B'.  The  work  done  in  stretching 
the  string  through  the  range  AB  is  now,  in  accordance  with  §  114,  repre- 
sented by  the  area  AA'B'B,  the  area  which  is  shaded  in  fig.  85. 


Fig.  85 


GRAPHICAL  REPRESENTATION  OF  WORK   151 


The  area  of  this  figure  is  clearly  equal  to  AJi  multiplied  by  the  ordinate 
at  the  middle  point  of  AB.  This  ordinate  represents  the  tension  of  the 
string  when  its  length  is  equal  to  l(OA  +  OB),  so  that  we  again  obtain  the 
result  of  §  113,  namely 

(work  done)  =  (range  of  stretching,  AB) 

X  (tension  at  halfway  stage  of  stretching). 

116.  The  indicator  diagram.  The  graphical  representation  of 
work  explained  in  §114  is  made  use  of  in  practical  engineering. 
Suppose  that  00'  is  the  distance  traveled  by  a  piston  inside  a 
cylinder.  When  the  piston  is  in  any  position  F,  let  the  pressure 
acting  on  the  piston  be  measured, 
and  let  a  line  PF'  be  drawn  at 
right  angles  to  00'  to  represent  it 
on  any  assigned  scale.  As  the 
piston  moves  along  the  range  00' 
and  then  back  along  the  range 
O'O,  the  point  F'  will  describe  a 
closed  curve  AF'BF"A,  which  is 
called  the  indicator  diayram  of  the 
motion  of  the  piston. 

The  work  done  by  the  steam  on 
the  piston  in  its  forward  motion  is, 
as  we  have  seen,  represented  by  the  area  AF'BO'FOA  inclosed  be- 
tween the  curve  AF'B  and  the  axis  00'.  This  work  is  expended  in 
movmg  the  piston  forward  in  opposition  to  the  thrust  in  the  piston 
rod.  Similarly  the  work  done  by  the  steam  on  the  piston  in  its  back- 
ward motion  is  represented  by  the  curve  B0'F0AF"B  mclosed  be- 
tween BF"A  and  the  axis  00' ,  this  area  being  taken  negatively,  since 
the  piston  is  now  moving  in  opposition  to  the  pressure  at  work  on  it. 

Thus  the  whole  work  done  on  the  piston  is  represented  by  the 
difference  of  tliese  two  areas,  and  this  is  easily  seen  to  be  the  area 
AF'BF"A  of  the  mdicator  diagram  itself.  Hence,  to  find  the  rate 
at  which  an  engme  is  performmg  %vork,  it  is  only  necessary  to 
measure  the  area  of  its  indicator  diagram  and  the  number  of 
revolutions  per  unit  time. 


-1. 

^ 

p' 

0 

r 

— > 

— > 

— > 

0' 


Fig.  8(5 


152  woiiK 

Work  done  against  Force  Oblique  to  Direction  of  Motion 

117.  So  far  we  have  only  considered  cases  in  which  the  force 
acts  in  a  direction  exactly  opposite  to  that  in  which  the  particle 
moves.  We  may,  however,  have  to  calculate  the  work  when  the 
motion  makes  any  angle  with  the  direction  of  the  force. 

When  a  body  is  moved  at  right  angles  to  the  force  acting  on  it, 
the  work  done  will  clearly  be  nil ;  e.g.  in  moving  a  weight  about 
on  a  horizontal  surface  no  work  is  done  against  gravity. 

We  can  now  find  the  amount  of  work  done  when  a  body  is  moved 

in  a  direction  making  any  angle  with  the  force  acting  on  it.    Let  a 

body  be  moved  from  F  to  Q,  a  small  distance  ds  of  its  path,  while 

iQ  acted  on  by  a  force  B,  of  which  the  line  of 

\^S(;^  action  makes  an  angle  (f)  with  QF.    Eesolve 


/ 1^- — '     c  F  into  two  components,  F  cos  <f>  along  QF 

and  F  sin  cb  perpendicular  to  QF.    The  work 
Fig.  87  ,  -in  -r.  •       i 

done  against  the  force  F  is  the  same  as  the 
work  which  would  be  done  if  these  two  forces  F  cos  cf),  F  sin  ^  were 
acting  on  the  body  simultaneously.  The  work  done  against  the 
former  force  would  be  F  cos  (f>  ds ;  that  against  the  latter  would 
be  nil.    Thus  the  whole  amount  of  work  done  is  F  ds  cos  (f). 

118.  Let  F  have  components  X,  Y,  Z,  and  let  the  element  of 
path  FQ  have  direction  cosines  /,  m,  n.    The  direction  cosines  of  the 

line  of  action  of  F  are 

X       Y      Z 

■ — i     - — >     — ) 
F      F      F 


and  since  this  makes  an  angle  tt  —  cf)  with  FQ,  we  must  have 

X  i"         ^ 

f' 


cos  (tt  —  ^)  =  / \-  m—^-\-  n- 

F  F 


Hence  F  ds  cos  (f)  =  —  ds  {IX  +  mY  +  nZ) 

=  —  {Xdx  +  Ydy  +  Zdz), 

where  dx,  dy,  dz  are  the  projections  of  ds  on  the  axes.  This  gives 
an  analytical  expression  for  the  work  done  m  a  small  displace- 
ment.   By  mtegration,  we  can  tind  the  work  done  in  any  motion. 


WORK  OF  RAISING  BODIES  AGAINST  GRAVITY     153 

119.  Work  of  raising  a  system  of  bodies  against  gravity.   If  a 

particle  of  mass  m  is  moved  a  distaneev/.s  along  a  path  making  an 
angle  (^  with  the  vertical  (upwards),  the  work  done  is  riiy  cos0  ch. 
Since  the  distance  through  which  the  particle  is  raised  is  ds  cos  0, 
we  may  say  that  the  work  done  is  equal  to  the  weight  of  the 
body  (mg)  multiplied  by  the  distance  through  which  the  particle 
is  raised. 

By  taking  the  particle  along  any  path,  and  adding  together  the 
amounts  of  work  done  on  the  successive  elements  of  the  path,  we 
find  that  the  total  work  done  against  gravity  is  erpial  to  the  weight 
of  the  particle  multiplied  by  the  total  vertical  distance  through 
which  the  body  has  been  raised. 

120.  Let  us  suppose  that  we  move  a  number  of  particles  of 
masses  m^,  m^,  •  •-.  Let  their  heights  above  the  ground  before  the 
motion  be  h^,  h^,---,  and  let  their  heights  at  the  end  of  the  motion 
be  h[,  111,  ••■.  The  work  done  against  gravity  on  the  first  particle 
is  liiiffi^  —  ^ii);  by  addition  of  such  quantities,  the  total  work 
done  against  gravity 

=  m,g  {]/[  -Ji,)  +  111. (J  (//',  -  A,)  +  ■  ■  • 

=  il(^"hK-^»h^'^-  (35) 

Now  let  M  be  the  total  mass  of  the  particles,  and  let  H,  //' 
denote  the  heights  of  the  center  of  gravity  of  all  the  particles 
above  the  ground  before  and  after  the  motion  respectively.  Then, 
by  the  formula  of  §  86,  we  have 

H  =  — — : =  =7: ' 

so  that  ^S,"h^h  ~  ^m, 

and,  similarly,  T^^'^^'i  =  j^UI'. 

Thus  the  total  work,  as  given  by  expression  (35),  becomes 
g{MH'  -  3IH)  =  Mg(J{'  -  JI). 


154  WORK 

Tlius  the  total  work  done  against  gravity  is  equal  to  the  total 
weight  of  the  particles  multiplied  by  the  vertical  height  through 
which  the  center  of  gravity  of  the  particles  has  been  raised. 

WOKK    PERFOEMED    AGAINST   A    CoUPLE 

121.  Theorem.  If  a  rigid  hody  acted  on  hy  a  system  of  forces 
he  given  any  small  rotation  through  an  angle  e  about  any  axis,  the 
ivorlc  done  is  Ge,  lohere  G  is  the  moment  about  this  axis  of  the 
forces  opposing  the  motion. 

Let  the  axis  of  rotation  be  supposed  to  be  a  line  perpendicular 
to  the  plane  of  the  paper,  meeting  it  in  the  point  L.    Let  a  typical 

force  be  a  force  F  acting  on  the  particle 
A  of  the  body. 

As  the  result  of  the  rotation,  let  A 

move  to  a  position  A',  so  that  the  angle 

ALA'  is  equal  to  e,  the  angle  through 

which  the  body  has  been  turned. 

Then,  durmg  the  rotation,  the  point 

Fig.  88  ^  ,..,,, 

of  application  of  the  force   F  moves 
from  A  to  A',  and,  therefore,  the  work  done 

=  F-AA'-  cos  (^, 

where  (f>  is  the  angle  between  F  and  AA', 

=  AA'  X  component  of  F  along  AA' 

=  €  X  LA  X  component  of  F  along  AA' 

=  e  X  moment  of  F  about  the  axis  of  rotation. 

If  the  rigid  body  is  acted  on  by  a  number  of  forces  applied  to 
its  different  particles,  we  find,  on  summation,  that  the  total  work 
done 

=  e  X  sum  of  the  moments  of  all  these  forces 

about  the  axis  of  rotation 
=  Ge,  where  G  is  the  moment  about  the  axis  of 
rotation  of  all  the  forces. 


YlliTLAL   WOKK  155 

EXAMPLES 

1.  A  man  who  weighs  140  pounds  walks  up  a  mountain  path  at  a  slope  of 
30  degrees  to  the  horizon  at  the  rate  of  1  mile  per  hour.  Find  his  rate  of  work- 
ing in  raising  his  own  weight  in  horse  power. 

2.  At  wliat  liorse  power  is  an  engine  working  which  liauls  a  train  of  1000 
tons  up  an  incline  of  1  in  200  at  12  miles  an  liour,  tlie  resistance  due  to  friction 
being  ^-J  ^  of  the  weight  of  the  train  ? 

3.  An  automobile  weigliing  1  ton  can  run  up  a  hill  of  1  in  60  at  8  miles  an 
hour.  Taking  the  resistance  due  to  friction  as  -^^  of  the  weight  of  the  car, 
find  at  what  rate  it  could  run  down  the  same  hill,  assuming  the  horse  power 
developed  by  the  engine  to  remain  tlie  same. 

4.  A  cargo  of  stone  weighing  18  tons  is  unloaded  from  a  barge  on  to  a  quay 
30  feet  above  tlie  barge  by  cranes  worked  by  an  engine.  If  the  unloading  takes 
three  hours,  find  tlie  average  horse  power  at  whicli  tlie  engine  has  been  working. 

5.  Assuming  that  a  man  in  walking  raises  his  center  of  gravity  through  a 
vertical  height  of  one  inch  at  eveiy  step,  find  at  what  horse  power  a  man 
is  working  in  walking  at  4  miles  an  hour,  his  stride  being  33  inches,  and  his 
weight  168  pounds. 

6.  A  cyclist  and  his  machine  weigh  200  pounds,  and  he  rides  up  an  incline 
of  1  in  80  at  15  miles  an  hour.  His  bicycle  is  geared  to  72  inches,  and  the 
length  of  his  cranks  is  7  inches.  Find  the  average  vertical  pressure  of  his  foot 
on  the  pedal,  assuming  this  pressure  to  exist  only  during  the  downward  motion 
of  the  pedal. 

7.  A  single-screw  ship  has  engines  of  5000  horse  power,  and,  when  working 
at  full  power,  the  engines  make  75  revolutions  per  minute.  Find  the  couple 
transmitted  by  the  shaft. 

8.  When  one  body  rolls  on  another,  there  Ls  found  to  be  a  couple  opposing 
the  motion,  equal  to  that  produced  by  the  normal  reaction  at  the  end  of  an 
arm  of  length  I,  where  I  is  called  the  coefficient  of  rolling  friction. 

If  a  railroad  truck  runs  on  wheel  of  radius  a,  show  that  the  resistance  to  its 
motion  produced  by  rolling  friction  is  l/a  times  its  weight. 

The  Principle  of  Virtual  Work 

122.  By  a  small  displacement  is  meant  for  the  present  a  motion 
in  which  each  particle  of  a  system  is  displaced  from  its  original 
position  through  a  distance  which  is  so  small  that  it  may  be  treated 
as  an  infinitesimal  quantity  of  which  the  square  may  be  neglected. 
If  the  system  is  under  the  action  of  forces,  work  will  be  done  in 
performing  any  small  disiAacement.  Since  the  displacement  is 
supposed  to  be  a  small  quantit}-,  the  work  performed  will  also  be 
a  small  quantity. 


156  WORK 

If  any  particle  is  in  equilibrium,  the  resultant  force  acting 
on  it  vanishes,  so  that  the  work  done  in  any  small  displacement 
of  the  particle  vanishes  to  a  higher  order  than  the  displacement. 
If  a  rigid  body,  or  system  of  rigid  bodies  or  particles,  is  in  equilib- 
rium, and  any  small  displacement  is  given  to  it,  the  work  done 
on  each  particle  is  nil,  so  that  the  aggregate  work  done  is  nil. 

123.  The  forces  acting  on  the  particles  of  the  system  may,  as 
in  v^  50,  be  divided  into  two  classes : 

(a)  forces  applied  to  the  bodies  from  outside ; 

(b)  pairs  of  actions  and  reactions  acting  between  the  particles 
of  the  bodies,  or  between  two  bodies  m  contact. 

In  calculatuig  the  work  done  in  a  small  displacement,  we  must 
take  account  of  the  work  done  against  all  the  forces  of  both  classes, 
but  shall  find  that  a  great  number  of  the  terms  arising  from  the 
forces  of  the  second  class  cut  one  another  out. 

124.  Let  us  first  consider  the  pair  of  forces  which  constitute 
the  action  and  reaction  between  two  particles  P,  Q  oi  a,  rigid  body. 
Let  the  amount  of  each  force  be  B,  its  direction  being  QF  or  FQ 

'  Q^   according  as  it  acts  on  P  or  Q.    Let 

the  effect  of  a  small  displacement  be 

to  move  P,  Q  to  P',  Q'  respectively,  and 

let  P'}),  Q'q  be  perpendiculars  drawni 

from  P',  Q'  to  PQ.    The  work  done 

against  the  force  E  acting  on  P  is  i?  X  P/-^,  while  that  done  against 

the  force  P  acting  onQis—Bx  Qq.    Thus  the  total  work  performed 

=  B{Pp—Qfi) 

=  BiPQ-pq) 

=  B{PQ  —  projection  of  P'Q'  on  P(J). 
Since  the  body  is  rigid,  the  length  P'Q'  is  equal  to  the  length 
PQ,  and  since  the  displacement  is,  by  hypothesis,  small,  the  angle 
between  P'Q'  and  PQ  is  small.  Thus  the  projection  of  P'Q'  on  PQ 
=  P'Q',  except  for  small  quantities  of  order  higher  than  the  first, 
=  BQ, 
so  that  the  work  performed  vanishes. 


P  J3 


YlIitUAL   WORK  107 

125.  Again,  the  work  performed  against  the  pair  of  forces  which 
constitute  action  and  reaction  between  two  smooth  surfaces  can 
be  seen  to  vanish. 

First  consider  the  case  in  which  one  body  is  held  at  rest  while 
the  second  is  made  to  slide  over  its  surface.  In  such  a  displace- 
ment the  work  performed,  if  any,  is  performed  agahist  the  reaction 
wliich  acts  on  the  moving  body.  Since  the 
force  acts  along  the  normal,  wliile  its  point  of 
application  necessarily  moves  in  the  tangent 
plane,  —  i.e.  at  right  angles  to  the  normal, — 
we  see  that  the  work  done  is  nil. 

The  most  general  motion  possible  for  the  two 
surfaces  is  compounded  of  a  motion  of  the  kind 
just  described  and  a  motion  in  which  the  two 
surfaces  move  as  a  rigid  body.  The  work  done 
in  the  first  part  of  the  displacement  has  just  been  seen  to  be  zero, 
the  work  done  in  the  second  part  of  the  displacement  vanishes  by 
§  124;  hence  the  total  work  vanishes,  proving  the  result  required 

126.  The  results  just  proved  are  not  true  if  the  contact  between  the 
surfaces  is  rough.  The  work  done  in  such  a  case  depends  on  the  magni- 
tude of  the  frictional  forces,  and  as  it  is  generally  as  difficult  to  determine 
the  amount  of  these  forces  as  to  solve  the  w^holg  problem,  the  method  of 
virtual  work  is  not  of  any  value  in  such  cases. 

127.  We  have  now  seen  that  a  large  number  of  forces  nuiy  be 
left  out  of  account  altogether  in  calculating  the  work  done  in  a 
small  displacement,  and  the  principle  of  virtual  work,  which  states 
that  when  a  system  is  in  equilibrium  the  work  done  in  any  small 
displacement  is  zero,  requires  us  only  to  calculate  the  work  jier- 
formed  against  external  forces,  and  not  that  performed  against  the 
mternal  actions  and  reactions  of  rigid  bodies. 

128.  Systems  of  pulleys.  An  important  application  of  the  prin- 
ciple of  virtual  work  is  the  following :  Let  us  suppose  that  we  have 
any  arrangement  of  pulleys  and  inextensible  ropes,  the  ropes  hav- 
ing two  free  ends,  —  to  one  of  which  the  weight  to  be  raised  is 
attached,  and  to  the  other  of  which  the  power  is  applied.    Let  these 


158 


WUEK 


two  free  ends  of  rope  be  called  the  wei<i,ht  end  and  the  power  end 
respectively,  and  let  us  suppose  that  the  arrangement  is  such  that, 
in  order  to  move  the  weight  end  tlirougli  1  mch,  the  power  end 
must  be  moved  tlirougli  n  inches.  Let  a  weight  W  be  attached  to 
the  weight  end,  and  let  us  suppose  that  it  is  found  that  a  force  P 
must  be  applied  to  the  power  end  to  maintain  equilibrium. 

We  now  have  forces  P  and  F  in  equilibrium.  To  find  the  rela- 
tion between  them,  let  us  give  the  system  a  small  displacement. 
Let  us  move  the  weight  W  a  distance  ds,  then,  if  the  rope  is  not  to 
be  stretched,  we  must  suppose  the  power  P  inoved  through  a  dis- 
tance n  ds.  The  work  done  by  external  force  consists  solely  of 
the  work  performed  on  the  power  end  of  the  rope,  namely  P  n  ds, 
and  the  work  performed  in  moving  the  weight  against  gravity, 
namely  Wds.  These  are  of  opposite  signs, — if  we  raise  the  weight, 
Wds  must  be  taken  positively  and  P  n  ds  negatively,  and  vice  versa. 
If  the  system  was  initially  in  equilibrium,  the  total  work  performed 
by  external  forces  in  this  small  displacement  must  vanish,  so  that 
the  equation  of  equilil)rium  is  seen  to  be 

Wds  —  Pnds  =  Q, 

W 

so  that  P  =  —  > 


giving  the  relation  between  power  and  weight. 
This  investigation  assumes  that  friction, 
etc.,  may  be  neglected,  and  also  neglects  the 
weight  of  the  movmg  ropes  and  pulleys. 

As  an  instance  of  a  system  of  pulleys,  let  us 

consider  the  arrangement  shown  in  fig.  91. 

Pjq   gj  There  are   two   blocks   of  pulleys,  A   and  B. 

The  former  is  fixed,  w'hile  the  latter  is  free  to 
move,  and  has  the  weight  W  suspended  from  it.  The  rope,  starting 
from  the  power  end,  passes  first  round  a  pulley  of  block  A,  then  round 
one  of  block  B,  then  round  one  of  block  A ,  and  so  on  any  number  of 
times,  until  finally  its  end  is  fastened  to  block  B.  To  find  the  relation 
between  P  and  W,  we  need  only  find  the  number  n.  Let  us  suppose  that 
in  addition  to  the  free  power  end  of  the  •  roi)e  the  number  of  vertical 
ropes  is  s.     Then,  if  we  pull  the  power  end  until  the  weight  end  is  raised 


VIKTUAL    WOllK 


159 


1  inch,  we  shall  shorten  each  of  these  .s  roj)es  by  1  inch,  and  so  lengthen 

the  power  end  by  s  inches.     Thus  n  =  s,  so  that,  in  this  case,  P  =  —  . 

s 
For  instance,  with  two  pulleys  in  the  lower  block  and  three  in  the 
upper  block  the  value  of  n  will  be  5,  so  that  each  pound  of  power  will 
support  5  pounds  of  weight,  —  a  man  pulling  with  a  vertical  pull  of 
100  pounds  could  support  a  weight  of  500  pounds,  and  as  soon  as  his  pull 
exceeds  100  pounds,  he  will  raise  the  weight  of  500  pounds. 


ILLUSTRATIVE  EXAMPLES 

1.  As  a  first  example  of  the  principle  of  virtual  work,  let  us  suppose  that  we 
have  an  endless  elastic  string  of  natural  length  a,  modulus  X,  placed  over  a 
sphere  of  radius  b,  and  allowed  to  stretch  under  gravity.  We  might,  of  course, 
find  the  amount  of  stretching  in  the  equilibrium 
position  by  resolving  forces,  but  we  can  get  it 
more  readily  by  the  method  of  virtual  work.  Let 
us  suppose  that,  when  in  equilibrium,  the  string 
lies  on  a  small  circle  of  angular  radius  6.  Let  a 
small  displacement  be  given,  this  consisting  of 
each  element  of  the  string  being  displaced  down 
the  surface  of  the  sphere,  so  that  the  string  forms 
a  new  circle  of  angular  radius  6  +  dd.  The  length 
of  the  string  when  forming  a  circle  of  angle  6  was 
2  irb  sin  0  ;  the  increase  in  this  when  d  is  changed 

to  e  +  dd  is  dd  —  (2  7rb  sin  6)  or  2  Trb  cos  0  dd.    The 

cd 
work  done  in  stretching  the  string  by  this  amount 
is  T ■2Trb  cos 6 dS,  where  T  is  the  tension.    Work 

is  also  done  against  (or,  in  this  particular  case,  with)  the  force  of  gravity.  The 
height  of  the  center  of  gravity  of  the  string  when  forming  a  circle  of  angle  0  is 
6  cos  ^  ;  on  increasing  ^  to  ^  +  dd,  this  increases  by  —  6  sin  edd,  so  that  the  work 
done  against  gravity  is  —lobsindde.  We  have  now  calculated  all  the  work 
performed  in  the  small  displacement;  by  the  principle  of  virtual  work,  the 
total  amount  of  this  work  must  be  nil,  so  that 

-  ivb  sin  ddd  +  T-2irb  cos  0d0  =  O. 


Fii;.  <.fJ 


Thus, 


—  tan  0, 
2ir 


and  the  length  of  the  strin<r  correspond iu!;  to  tension  T  is,  as  we  have  seen, 

a(l  +  -^^  tan  ^ )  =  2  7r6  sin  6, 
\        2  7r\  / 


Hence 
an  equation  giving  0. 


IGU  WORK 

2.  Gearing  of  a  bicycle.  As  a  second  example,  let  us  apply  the  principle  of 
virtual  work  to  the  mechanism  of  a  bicycle.  Let  the  length  of  the  crank  be  a, 
and  let  the  bicycle  be  geared  to  b  inches,  so  that  each  revolution  of  the  pedals 
causes  the  machine  to  move  as  far  forward  as  it  would  in  one  revolution  of  a 
wheel  of  b  inches  diameter.  Let  us  find  what  pressure  must  be  exerted  on  the 
pedal  by  a  rider  in  order  that  the  machine  may  move  forward  against  an 
opposing  frictional  force  of  w  pounds  weight. 

Let  us  give  the  machine  a  small  displacement,  the  cranks  being  supposed  to 
turn  through  an  infinitesimal  angle  e,  and  the  wheels  and  machine  moving 
forward  accordingly.  Since  the  gearing  is  to  b  inches,  the  distance  moved  by 
the  machine  as  a  whole  will  be  i  be  inches,  while  the  distance  moved  by  the  pedal, 
taking  the  machine  itself  as  frame  of  reference,  will  be  ae.  Let  TF" pounds  weight 
be  the  force  exerted  on  the  pedal  when  the  machine  is  just  on  the  point  of  motion, 
so  that  the  machine  is  in  equilibrium  under  this  force  acting  on  the  pedal,  and 
the  backward  pull  of  w  pounds  due  to  friction.  The  equation  of  virtual  work  is 
W  ■  ae  —  w  ■  h  be  =  0, 

so  that  the  required  force  is  TF  =  —  w. 

%a 

Thus  the  force  is  directly  proportional  to  the  gearing  of  the  machine,  but 
inversely  proportional  to  the  length  of  the  cranks. 

3.  Four  rods  of  equal  weight  w  and  length  a  are  freely  jointed  so  as  to  form  a 
rhombus  ABCD.  The  framework  stands  on  a  horizontal  table  so  that  CA  is  vertical, 
and  the  whole  is  prevented  from  collapsing  by  a  weightless  ineztensible  string  of 

length  I  ivhlch  connects  the  points  B,  D.    It  is 

^^,/f.^  required  to  find  the  tension  in  this  string. 

\.  rt  To  find  the  tension  by  the  principle  of 

^V,^  virtual  work,  we  must  of  course  find  a  small 

^^B  displacement  such  that  work  is  done  in  oppo- 

^/^        sition  to  the  tension,  or  otherwise  the  tension 

y^  would  not  enter  into  the  equations  at  all. 

^  Since   the   string   is   inextensible,  it   is   not 

possible  in  actual  fact  to  stretch  it  and  so 
Fig.  93  ^ 

perform  work  against  its  tension.    We  can 

however  imagine  it  to  be  stretched  in  spite  of  its  actual  inextensibility,  or, 

what  comes  to  the  same  thing,  we  can  imagine  it  replaced  by  an  extensible 

string  of  the  same  length  and  having  the  same  tension.     It  is  now  easy  to 

arrange  a  displacement  of  the  kind  required. 

Let  us  imagine  that  the  framework  is  displaced  in  such  a  way  that  A  moves 

vertically  downwards  towards  C,  while  C  remains  at  rest.    Let  the  displacement 

be  such  that  the  angle  HAC  is  increased  from  0  to  ^  +  dQ.    The  length  I  of  the 

string  which  corresponds  to  the  angle  6  is  given  by 

J  =  2  a  sin  ^, 

from  which,  by  differentiation,  we  obtain 

dl  -2a  cos  d  d9. 


ILLUSTRATIVE  EXAMPLES 


161 


giving  tlie  relation  between  the  increments  dl,  dd  in  I  and  0.  The  work  done 
against  the  tension  of  the  string  (T)  in  this  displacement  is  Tdl.  The  height 
of  the  center  of  gravity  of  the  whole  figure  above  C  is  initially  ^  AC,  or  a  cos  d, 
so  that,  as  in  §  120,  the  work  done  against  gravity  is 

4 10  dia  cos,  ff). 
Thus  the  total  work  jDerformed  by  external  forces  in  the  displacement  is 
Awd{aQ09,e)  +  Tdl, 
or,  on  substituting  the  values  of  dl  and  d{a  cos  6), 

—  4  loa  sin  Odd  +  T-2a  cos  0  d0. 
For  equilibrium  this  must  vanish.    "We  must  therefore  have 

T=2w  tan d, 
giving  the  required  tension. 

4.  A  rod  of  length  I  and  weight  w  is  suspended  by  its  two  ends  from  two  points 
at  the  same  height  and  distant  I  apart,  by  two  strings  each  of  length  a.  Find  the 
couple  required  to  hold  the  rod  in  a  position 
in  which  it  makes  an  angle  6  with  its  equi- 
librium position. 

In  equilibrium  the  strings  are  vertical, 
the  two  ends  A,  B  oi  the  rod  lying  exactly 
underneath  the  two  points  of  suspension 

P,Q- 

As  the  rod  is  turned  from  its  equilibrium 
position,  we  c'an  imagine  its  middle  point  to 
rise  gradually  along  the  vertical  line  through 
the  original  position  of  this  middle  point. 
When  the  rod  has  been  turned  through  any 
angle  6,  let  the  height  through  which  this 
point  has  risen  be  x. 

Then  the  projection  of  the  length  PA' 
on  a  vertical  line  will  be  a  —  x,  while  its  projection  on  a  horizontal  plane, 

being  equal  to  the  horizontal  projection  of  AA',  will  clearly  be  Zsin-- 

Thus,  expressing  that  the  length  of  the  displaced  string  PA'  remains  equal 
to  its  original  value  a,  we  have 

a"^  =  {a  —  xy-  +  /-  sin-  -  •  (a) 

To  find  the  couple  retjuired  to  hold  the  rod  ai  an  angle  6,  let  us  suppose  that 
the  rod  is  held  in  eiiuilibrium  in  this  position  by  a  couple  G,  and  that  a  snuill 
displacement  occurs  in  which  6  is  increased  to  0  +  dd.  The  work  done  against 
the  couple  is  equal,  by  §  121,  to  —  Gdd,  the  negative  sign  being  taken,  since 
the  couple  aids,  instead  of  opposing,  the  motion.  The  work  done  against  gravity 
is  equal  to  w  dx.    Thus  the  equation  of  eijuilibrium  is 

-  Gde  +  w  dx  =  0. 


1G2  WOEK 

To  ubtain  the  relation  between  dd  and  dx  we  differentiate  equation  (a), 
obtaining 

—  '2ia  —  x)  dx  +  I-  sin  -  cos    dd  =  0. 
^  2        2 

Tims  G  =  IV  — 

dd 

ivl-  sni  -  cos  - 
_  2        2 

2{a  —  x) 
_         wl~  sine 

',  / .   ,, . ./ 

\  2 

giving  the  couple  required. 

EXAMPLES 


1.  A  square  ABCD  is  formed  by  joining  four  equal  rods  by  freely  moving 
hinges.  The  points  A,  C  are  joined  by  an  elastic  string  of  natural  length  equal 
to  a  diagonal  of  the  square,  and  of  modulus  \.  What  forces  must  be  applied 
to  the  points  B,  D  to  stretch  the  string  to  double  its  length  ? 

2.  Three  spheres  each  of  radius  a  and  weight  w  are  tied  to  a  point  P  by 
strings  of  natural  length  I  and  modulus  X,  and  hang  freely,  touching  each  other. 
Find  the  depth  of  their  centers  below  P. 

3.  The  mechanism  by  which  a  Japanese  umbrella  is  opened  is  such  that  each 
rib  tui-ns  through  an  angle  of  5°  for  every  inch  that  the  sliding  piece  is  moved 
up  the  stick.  If  there  are  18  ribs,  each  of  weight  i  ounce,  and  having  their 
centers  of  gravity  10  inches  from  their  piyots,  find  with  what  force  the  sliding 
piece  must  be  pushed  up  the  stick  to  open  the  umbrella,  when  the  stick  is  held 
vertically,  and  the  ribs  are  inclined  at  an  angle  of  30°  to  it. 

4.  The  hands  of  a  clock  are  balanced  with  countei-poises,  so  as  to  be  in 
equilibrium  in  any  position.  When  the  time  indicated  by  the  clock  is  5.10,  a 
bird  of  weight  w  suspends  itself  from  a  point  on"  the  minute  hand  which  is  six 
feet  from  the  pivot.  How  large  a  vertical  thrust  must  be  applied  to  the  hour 
hand,  also  at  a  point  six  feet  from  the  pivot,  to  restore  equilibrium  ? 

5.  A  clock  is  wound  by  raising  a  weight  of  20  pounds  through  a  distance  of 
3  feet,  this  enabling  the  clock  to  run  for  30  hours.  The  pendulum  and  escape- 
ment are  removed,  so  that  the  hands  will  "race"  unless  held  fast.  How  large 
a  couple  must  be  applied  to  the  minute  hand  to  prevent  this  occurring  ? 

6.  The  coupling  between  two  English  railway  carriages  consists  of  a  rod  with 
a  right-handed  and  a  left-handed  screw  cut  at  its  opposite  ends  and  turning 
in  nuts  attached  to  the  carriages.  If  the  pitch  of  each  screw  is  one  inch,  and 
the  rod  is  turned  by  a  force  of  56  pounds  acting  at  best  advantage  at  the 
end  of  a  lever  15  inches  long,  find  the  force  by  which  the  carriages  are 
drawn  together. 


POTENTIAL  ENERGY  163 

Potential  Energy 

129.  It  will  have  been  noticed  that  we  are  concerned  with  two 
different  kinds  of  work.  The  first  is  typified  by  the  work  done  in 
raising  a  weight  in  opposition  to  gravity,  tlie  second  by  the  work 
done  against  friction  in  hauling  a  train  along  a  level  road.  Tlie 
essential  difference  between  the  two  kinds  is  that  work  of  the  first 
kind  can  be  recovered  from  the  system  of  bodies  by  making  these 
bodies  themselves  perform  mechanical  work,  whereas  work  of  the 
second  kind,  when  once  expended,  can  never  be  regained.  In  rais- 
ing a  weight  we  may  be  said  to  be  storing  up  work  rather  than 
spending  it,  for  the  weight  can  at  any  time  be  made  to  yiekl  back 
all  the  work  devoted  to  raising  it.  If  we  raise  a  weiglit  -w  tb  rough 
a  distance  h,  the  work  done  on  the  weight  is  wh ;  on  letting  the 
weight  descend  to  its  original  j)osition,  the  work  done  for  us  Ijy 
the  weight  is  ivh,  so  that  the  total  work  performed  on  the  weight 
is  nil. 

On  the  other  hand,  in  haiding  a  mass  a  distance  s  against 
a  frictional  force  F,  the  work  performed  is  Fs.  To  bring  the 
mass  back  to  its  original  position,  we  have  to  expend  an  addi- 
tional amount  of  work  Fs,  so  that  the  total  work  performed  is 
2  Fs.  This  brings  out  the  essential  difference  between  the  two 
types  of  work  and  between  the  two  systems  of  forces  against 
which  the  work  is  performed. 

130.  Definition.  When  the  forecs  acting  on  a  sy stein  of  bodies 
are  of  such  a  nature  that  the  alijehraic  total  ivork  done  in  jJ^'form- 
ing  any  series  of  displaxements  tvhicli  bring  the  system  back  to  its 
original  configuration  is  nil,  the  system  of  forces  is  said  to  be  a 
conservative  system. 

The  algebraic  work  being  nil.  the  work  done  on  the  system  in  taking 
it  to  any  confignration  is  equal  and  opposite  in  sign  to  the  work  done  on 
the  system  in  allowing  it  to  resnme  its  former  contiguration.  so  that  all 
the  work  spent  can  be  regained.  The  work  is  accordingly  stored  np,  or 
conserved. 


164  WORK 

A  small  amount  of  reilection  will  show  that  a  system  of  forces 
is  conservative  if  the  only  forces  which  come  into  play  are  some 
or  all  of  the  following : 

(a)  gravity; 

(6)  reactions  in  wliich  the  contact  is  perfectly  smooth ; 

(c)  tensions  of  strings,  extensible  or  inextensible. 

On  the  other  hand,  if  any  one  or  more  forces  of  the  following 
types  come  into  play  (so  that  work  is  performed  against  them), 
the  system  of  forces  is  non-conservative  : 

(a)  reactions  in  which  the  contact  is  rough ; 

(h)  resistance  of  the  air. 

131.  Theorem.  The  work  done  on  a  system  of  bodies  acted  on 
1)7/  conservative  forces,  in  tnoving  from  one  configuration  P  to  a 

second  configuration  Q,  is  inde2:)endent  of  the  scries  of 
configurations  through  wloich  the  system  moves  in 
passing  from  P  to  Q. 

To  prove  this,  let  us  denote  the  work  done  in  pass- 
ing from  P  to  Q  through  one  series  of  configurations 
by  W^,  that  done  in  passing  through  any  second  series 
by  IF2,  and  that  done  in  returning  from  Q  to  Phj  any 
third  series  of  configurations  by  W^.  If  we  pass  from 
P  to  Q  by  the  first  series  and  back  from  Q  to  P  hj 
the  third,  the  total  work  done  is  nil,  so  that 

T[;  +  Tr;  =  0. 

So,  also,  if  we  pass  from  P  to  Q  by  series  2  and  back  from  Q  to 

P  by  series  3, 

w,  +  W,  =  0. 

Thus  1J\  =  W^ ,  which  proves  the  theorem. 

132.  Definition.  Talcing  any  configuration  P  as  standard,  the 
work  done  in  moving  a  system  of  bodies  from  the  configuration 
P  to  the  configuration  Q  is  spoken  of  as  the  potential  energy  of 
configuration  Q. 


POTENTIAL   ENERGY  165 

The  potential  energy,  accordingh',  iueasures  the  work  which 
has  been  stored  up  in  placing  the  system  in  configuration  Q. 

Theorem.  The  worli  done  in  moving  a  system  from  a  config- 
uration (1)  to  a  second  configuration  (2)  against  conservative  forces 
is  W„  — Tf\,  where  XF^,  W.-,  are  respectively  the  potential  energies  in 
configurations  (1)  and  (2). 

For  if  P  is  the  standartl  configuration,  the  work  from  P  to  (1) 
is  T^Fj ;  the  work  from  P  to  (1)  plus  that  from  (1)  to  (2)  is  W„,  so 
that  the  work  from  (1)  to  (2)  is  W^—W\. 

133.  Theorem.  If  a  system  of  bodies  is  in  a  configuration  of 
potential  energy  W,  and  if  x,  y,  z  ure  the  coordinates  of  any 
particle,  the  resultant  force  acting  on  the  particle  has  components 

cW  cW  cW 

— , — ,     —  _— — . 

C£  cy  cz 

To  prove  this,  let  us  imagine  that  we  give  the  system  a  small 
displacement,  which  consists  in  moving  the  single  particle  at  x,  y,  z 
a  distance  dx  parallel  to  the  axis  of  x.  If  X,  Y,  Z  are  the  compo- 
nents of  the  force  acting  on  it,  the  w^ork  we  do  in  the  displace- 
ment is,  as  in  §  118,  equal  to  —  X  dx.    This  work  is  also  equal  to 

dW 
the  increase  in  the  potential  energy,  namely  -77-  dx,  so  that  we  have 

—  Xdx  =  —  dx. 

dx 

OTJT' 

Thus  X= ^—>  and  similarly  we  may  prove  that 

dx 

dy  cz 

134.  Theorem.  If  a  system  of  bodies  is  in  a  configuration  of 
potential  energy  W,  and  if  6  is  an  angle  giving  the  orientation  of 
a  rigid  body  of  the  system  about  any  line,  the  moment  about  this  line 
of  the  forces  acting  on  the  rigid  body  {reckoned  positive  if  tending 
to  rotate  it  in  the  direction  of  6  increasing)  is 

_cW 
86' 


166  WORK 

For,  let  us  give  the  system  a  small  displacement,  which  consists 
in  turning  the  body  in  question  through  a  further  angle  dd  about 
the  selected  line,  so  that  6  becomes  changed  mto  6  +  dd.    The 

cW 
increase  in  potential  energy  is  i^dd,  while  the  work  performed 

Cu 

is,  by  the  theorem  of  §  121,  equal  to  —G  dO,  where  G  is  the  moment 
about  the  axis  of  all  the  forces  acting  on  the  rigid  body. 

Thus  ^-^de  =  -Gdd, 

CO 

dW 
so  that  G  =  — — r> 

oa 

the  result  required. 

135.  Theorem.  In  a  position  of  equilibrium  of  a  system  of 
bodies,  the  potential  energy  W  is  cither  a  maximum  or  a  minimum,. 

The  potential  energy  is  a  function  of  all  the  coordinates  of  all 
the  particles  of  which  the  system  of  bodies  is  composed,  say 

"^1 '  2^1 '  '^1 5    "^2 '  ^2  >  ~2  j   etc. 
If  the  position  is  one  of  equilibrium,  each  particle  is  in  equilib- 
rium, so  that  the  components  of  the  forces  acting  on  each  particle 
vanish  separately  by  §  33.    By  §  133  the  condition  for  this  is 

dx^  dy^  dz^ 

-^—  =  0,  etc. 

cx^ 

But  these  are  exactly  the  conditions  that  W  shall  be  a  maximum 
or  a  minimum. 

136.  The  converse  of  this  theorem  is  also  true. 

Theorem.  //  the  potential  energy  of  a  system,  of  bodies  is  either 
a  maximum  or  a  minimum  in  any  configuration,  then  the  con- 
figuration is  one  of  equilibrium. 

For,  with  the  notation  of  the  previous  section,  if  W  is  a  maxi- 
mum or  a  minimum,  it  follows  that 

—  =C,      —-Q      — -0 

dx^         '      ^Vi         '      ^^1 


POTENTIAL  ENERGY  167 

a-  ^^f'  ^W'  dJV  ^-  .         .      ,        . 

bince  —- — ,  ~- — , ^—  are  the   components  of  the  force 

cx^         cy^  cz^ 

acting  on  particle  (1),  these  equations  indicate  that  particle  (1) 
is  in  equilibrium.  Similarly,  it  follows  that  the  other  ])articles 
are  in  equilibrium,  giving  the  result. 

137.  An  important  special  case  of  these  theorems  arises  when 
the  only  forces  which  do  any  work  in  a  displacement  are  the 
weights  of  the  bodies  of  which  the  system  is  composed.  If  M  is 
the  mass  of  the  whole  system,  and  if  h  is  the  height  of  its  center 
of  gravity  above  any  standard  horizontal  plane,  the  potential 
energy  is,  by  §  120,  Mgli,  and  this  is  a  maximum  or  a  minimum 
when  h  is  a  maximum  or  a  minimum.    Thus  we  have  the  thecjrem  : 

In  a  system  of  bodies  in  ivhich  the  only  forces  luhich  perform 
work  in  a  displacement  are  those  of  gravity,  the  configurations  of 
equilibrium  are  those  in  which  the  height  of  the  center  of  gravity 
is  a  maximum  or  a  minimum. 


EXAMPLES 

1.  Two  uniform  rods,  each  of  length  I,  are  freely  jointed  at  their  extremities 
and  placed  over  a  smooth  cylinder  of  radius  a  of  which  the  axis  is  horizontal. 
Find  the  angle  which  the  rods  make  with  the  horizontal  when  in  ecjuiiibrium. 

2.  An  elliptic  disk  is  weighted  so  that  its  center  of  gravity  is  halfway 
between  its  center  and  one   extremity  of  its  major  axis.     8how  that  if  its 

eccentricity  is  greater  than  there  will  be  four  positions  of  equilibrium  in 

V2 
which  the  disk  stands  vertical  on  a  horizontal  plane,  but  otherwise  only  two. 

3.  A  horizontal  rod  of  weight  W  has  its  center  pierced  by  a  fixed  vertical 
screw  on  which  it  turns,  one  revolution  raising  or  lowering  it  by  ]  inch.  If 
there  is  no  friction,  find  the  couple  required  to  hold  it  at  rest. 

4.  A  plug  of  weight  W  is  made  in  the  shape  of  a  pyramid  of  square  cross 
section.  It  is  placed  with  its  axis  vertical  in  a  square  hole  of  side  c,  the  depth 
of  its  vertex  in  this  position  being  d  below  the  plane  of  contact.  Find  the  couple 
required  to  hold  it  turned  through  an  angle  6  and  still  having  its  axis  vertical. 

5.  A  smooth  parabolic  wire  is  placed  with  its  axis  vertical.  Two  beads  are 
strung  on  it,  and  are  connected  by  a  string  which  passes  through  a  smooth  ring 
at  the  focus.    Show  that  there  are  an  infinite  number  of  positions  of  etjuilibrium. 

6.  A  smooth  bowl  in  the  shape  of  an  ellipsoid  of  semi-axes  a,  6,  c  has  one  axis 
vertical.  Find  the  couple  required  to  hold  a  rod  of  length  I  in  a  horizontal  posi- 
tion in  the  bowl,  making  an  angle  d  with  a  position  of  equilibrium. 


168  WORK 

Kinetic  Energy 

138.  Suppose  that  a  moving  particle  is  acted  on  by  a  force 
of  which  the  direction  is  opposite  to  that  of  the  motion  of  the 
particle.  The  effect  of  the  force,  according  to  the  second  law  of 
motion,  is  to  produce  a  retardation  m  the  velocity  of  the  particle. 
The  velocity  of  the  particle  decreases  so  long  as  the  force  acts,  so 
that  if  the  force  continues  to  act  for  a  sufficient  time,  the  particle 
must  ultimately  be  reduced  to  rest. 

Consider,  for  example,  a  hammer  striking  a  nail.  The  reaction  between 
the  hammer  and  nail  is  a  force  in  the  direction  opposite  to  that  of  the 
motion  of  the  hammer,  and  this  ultimately  brings  the  hammer  to  rest. 
Again,  when  a  j^article  is  projected  vertically  upwards,  its  weight  after  a 
time  reduces  it  to  rest,  after  which  of  course  it  falls  back  to  the  ground. 

By  the  time  that  the  moving  body  has  been  reduced  to  rest 
the  point  of  application  of  the  force,  wliicli  has  moved  with  the 
moving  body,  has  moved  through  a  certain  distance.  Thus  a  cer- 
tain amount  of  work  has  been  done  by  the  moving  body.  We  are 
thus  led  to  the  conception  of  the  motion  of  a  body  possessing  a 
capacity  for  doing  work. 

For  instance,  in  the  previous  examples,  the  motion  of  the  hammer  has 
driven  the  nail  into  position,  and  the  motion  of  the  particle  projected  into 
the  air  has  raised  it  to  a  certain  height  above  the  earth's  sm'face. 

139.  Let  us  suppose  that  a  particle  moving  with  velocity  v  is 
opposed  by  a  force  P  (in  absolute  vmits)  acting  m  the  direction 
opposite  to  that  of  the  motion  of  the  particle.  Let  the  particle 
descrilje  a  distance  ds  in  opposition  to  this  force  m  time  dt,  and 
let  its  velocity  change  from  v  to  v  —  dv  in  this  time.    The  particle 

then  has  a  retardation  — -  in  the  direction  of  its  motion,  or,  what  is 

•  dt  7 

the  same  thing,  an  acceleration  —  in   the  direction  in  which  P 
*  dt 

is  actmg,  so  that  by  the  second  law  of  motion 

dv 
P  =  m  — 
dt 


KINETIC   ENEKGY  109 

The  work  done  by  the  paiticle  in  moving  the  distance  ds  in 
opposition  to  the  force  P  is 

Pds  =  VI  — ds, 
dt 

ds 
or,  since  —  is  the  same  as  the  velocity  v  of  the  particle, 

P  ds  =  niv  dv. 

Integrating,  we  find  that  the  whole  w^ork  done  by  the  particle 
before  being  reduced  to  rest  is 


/ 


Fds  =  lmv\  (36) 

Since  P  has  to  be  measured  m  absolute  units  (cf.  §  22),  it 
follows  (§  111)  that  the  work  ^mv^  will  also  be  measured  in 
absolute  units. 

Thus  whatever  the  magnitude  of  the  force  opposing  the  motion 
of  a  particle,  the  work  performed  by  the  particle  before  being 
reduced  to  rest  is  the  same,  namely  ^  mv'^  absolute  vmits  of  work. 

The  quantity  \7nv^  [measured  in  absolute  units)  is  called  the  kinetic 
energy  of  a  moving  particle.  It  is  equal  to  the  amount  of  work  which 
can  he  performed  by  the  p>article  before  being  reduced  to  rest. 

Siippose,  for  instance,  that  the  resistance  offered  by  a  nail  to  being 
driven  into  a  board  is  equal  to  the  weight  of  5000  pounds,  i.e.  that  it  would 
require  a  weight  of  5000  pounds  to  press  it  into  the  board.  Suppose  that 
it  is  driven  into  the  board  by  being  struck  with  a  hammer,  of  which  the 
head  weighs  10  pounds,  and  hits  the  nail  with  a  velocity  of  50  feet  per 
second.  Let  .s  be  the  distance  the  nail  is  driven  in  at  each  stroke  measured 
in  feet,  then  the  work  done  by  the  hammer  at  each  stroke  is  that  of  moving 
a  force  of  5000  poiinds  weight  —  or  5000  X  g  poundals — through  a-distance 
of  s  feet.  It  is  therefore  equal  to  5000  fjs  foot  poundals.  The  kinetic  energy 
of  the  hammer  in  striking  the  nail  is 

|wH-2=  4  .  10-50--^  =  12,500 
in  absolute  foot-pound-second  units.    Thus  from  the  relation  (oQ)  we  have 

the  equation  ■  ,  ^  _„„ 

^  5000  f/s  =  12,o00, 

in  which,  since  the  units  are  foot-pound-second  units,  we  may  take  g  =  32, 

and  so  obtain  .  , .  .     , 

s  =  aVs  feet  =  }g  inches. 


170  WOKK 

140.  Theorem.  During  the  motion  of  a  'particle  under  any  sys- 
tem of  forces,  the  increase  in  kinetic  ciiergy  is  eqiial  to  the  total 
work  done  on  the  particle  hy  external  agencies. 

Let  us  consider  motion  of  a  particle  from  one  position  P  to  a 
second  position  Q,  and  let  the  velocities  of  the  particle  at  these 
two  points  be  Vp,  Vq  respectively. 

Let  us  examine  the  motion  over  any  element  ds  of  the  path, 
and  let  the  velocities  at  the  beguuiing  and  end  of  this  element  be 
V  and  V  +  dv.  Let  P  be  the  force,  or  component  of  force  along  ds, 
which  acts  on  the  particle  while  it  describes  the  element  ds  of  its 
path.    If  dt  is  the  time  taken  to  describe  this  element  of  path,  the 

dv  . 

acceleration  is  — >  and  since  the  force  actuig  in  the  direction  of 

dt  ^ 

motion  is  P,  we  have,  by  the  second  law  of  motion, 

P  =  m-— • 
dt 

dv 
Hence,  as  in  §  139,     Pds  =  m  — -  ds 

^  dt 

ds  , 
=  7n  —  av 
dt 

=  mv  dv. 
Integrating  over  the  whole  path  from  P  to  Q,  we  obtain 


P  ds  =  m  i    V  dv 


=  I  m  v|  —  1-  m  Vp  (37) 

=  increase  in  kinetic  energy. 

The  left-hand  side  of  this  equation  represents  the  work  done  on 
the  particle,  proving  the  result  required. 

141.  The  work  performed  on  the  particle  by  external  forces  may 
be  regarded  also  as  equal  to  minus  the  work  performed  by  the  parti- 
cle on  external  agencies.  For  if  P  is  the  force  acting  on  the  particle 
along  ds,  it  follows  from  the  equality  of  action  and  reaction  that 
the  force  acting  on  the  external  agencies  from  the  particle  is  —  P, 


CONSEEVATION  OF  ENERGY  171 

so  that  the  total  work  performed  by  the  particle  is  —  P  ds.    Tlius 
the  theorem  can  be  stated  in  the  following  alternative  form : 

During  the  motion  of  a  particle  imder  any  system  of  forces,  the 
decrease  in  kinetic  energy  is  equal  to  the  total  work  done  hy  the 
particle  against  external  agencies. 

142.  If  the  system  of  forces  acting  on  the  particle  is  a  conserv- 
ative system,  the  value  of  —  /    Fds,  the  total  work  performed  by 

the  particle  on  external  agencies,  is  equal,  by  §  132,  to  ^V^J  —  Wp. 
Thus  equation  (37)  becomes 

or  again  TJ^  +  ^  m  v^  z=Wp+\  m  vf.,  (38) 

so  that  the  sum  of  the  potential  and  kinetic  energies  is  the  same 
at  Q  as  at  P,  provmg  the  theorem. 

The  sum  of  the  potential  and  kinetic  energies  is  called  the  total 
energy  of  the  particle. 

Conservation  of  Energy 

143.  The  kinetic  energy  of  a  system  of  bodies  is  obviously  equal 
to  the  sum  of  the  kmetic  energies  of  the  separate  particles.  The 
potential  energy  of  the  system,  as  has  been  seen,  is  the  sum  of 
the  potential  energies  of  its  particles. 

Thus  the  total  energy  of  a  system  is  equal  to  the  sum  of  tlie 
total  energies  of  the  separate  particles.  Since  the  total  energy  of 
each  particle  remains  constant,  it  follows  that  the  total  energy  of 
the  system  remains  constant. 

The  fact  that  the  total  energy  remains  constant  is  spoken  of  as 
the  Conservation  of  Energy.  An  equation  expressing  that  the  total 
energy  at  one  instant  is  equal  to  that  at  any  other  instant  is 
spoken  of  as  an  equation  of  energy. 

144.  As  an  illustration,  let  us  consider  the  firing-  of  a  stone  from  a 
catapult. 

Work  is  performed  in  the  first  place  in  stretching  the  elastic  of  the 
catapult,  and  the  work  is  stored  as  potential  energy  of  the  stretched  elastic. 
As  soon  as  the  catapult  is  released,  the  stone  is  acted  on  by  the  tension  of 


172 


WORK 


the  elastic  ;  the  stone  moves  under  the  accelerating  influence  of  this  tension, 
and  the  tension  of  the  elastic  slackens.  While  this  is  in  progress  the  stone 
is  acquiring  kinetic  energy,  while  the  stretched  elastic  is  losing  potential 
energy.  By  the  theorem  just  proved,  the  kinetic  energy  gained  by  the 
stone  must  be  just  equal  to  the  potential  energy  lost  by  the  elastic. 

When  the  stone  escapes  from  the  catapult,  most  of  the  potential  energy 
of  the  elastic  will  have  disappeared,  having  been  transformed  into  the 
kinetic  energy  of  the  stone.  After  this  a  further  transformation  of  energy 
may  take  place  while  the  stone  is  in  motion.  If  the  stone  moves  upwards, 
its  potential  energy  wall  increase,  so  that  there  must  be  a  corresponding 
decrease  in  its  kinetic  energy  —  its  speed  must  slacken.  On  the  other 
hand,  if  the  stone  moves  downwards,  the  potential  energy  will  decrease, 
so  that  its  kinetic  energy  will  increase  —  it  will  gain  in  velocity. 

145.  A  very  important  deduction  from  the  principle  of  the  con- 
servation of  energy  is  the  foUowmg : 

Theoeem.  If  a  ■particle  slide  along  any  smooth  curve,  being  acted 
on  hy  no  forces  except  gravity  and  the  reaction  with  the  curve,  and 
if  u,  V  he  the  velocities  at  two  p)oints  P,  Q  of  its  path,  then 

v'^u^^-'lgh,  (39) 

ivhere  h  is  the  vertical  distance  of  Q  helow  P,  —  i.e.  is  the  vertical 
projection  of  the  path  PQ  described  by  the  particle. 

Let  hp,  h^  denote  the  heights  of  P  and  Q  above  any  horizontal 
p^^^"      plane  —  for  instance,  the  earth's  surface. 
Then  when  the  particle  is  at  P  its  kinetic 
energy  is  ^  mu^,  and  its  potential  energy 
is  mghp.    Thus  its  total  energy  is 

'^mu'^  +  mghp. 

Similarly  at  Q  its  total  energy  is 

1-  mv"^  +  mgh(^. 

^^'  "'^  Since  the  system  of  forces  acting  is  a 

conservative  system,  the  total  energy  remains  unaltered.    Thus 

i  miv^  +  7nghp  =  ^  mv^  +  mgh^,, 
so  that  v^  —  %i?  —'Ig  {hp  —  h^)  =  2  gh, 

proving  the  theorem. 


CONSERVATION  OF  ENERGY  173 

146.  The  theorem  of  §  145  is  clearly  true  when  the  particle  is 
ascending,  in  which  case  h  is  negative  —  or  if  the  j^article  ascends 
during  part  of  its  path  and  descends  during  the  remainder.  ]\Iore- 
over,  the  particle  may  move  under  any  conservative  system  of 
forces,  provided  only  that  the  whole  potential  energy  arises  from 
the  weight  of  the  particle,  and  the  theorem  remains  true. 

It  is  true,  for  instance,  of  a  particle  tied  to  an  inextensible 
string,  or  of  a  particle  moving  freely  in  a  vacuum. 

To  illustrate  the  use  of  the  theorem,  let  us  suppose  tliat  a  bicyclist, 
riding  with  a  velocity  of  15  miles  an  hour,  comes  to  the  top  of  a  hill  of 
height  60  feet,  down  which  he  coasts.  Let  us  find  his  velocity  at  the 
bottom,  on  the  supposition  that  friction,  air  resistance,  etc.,  may  be 
neglected. 

Taking  the  top  and  bottom  of  the  hill  to  be  the  points  P,  Q  respectively 
of  the  theorem  just  proved,  we  have,  from  the  data  of  the  problem, 

^  =  60  feet, 

u  =  15  miles  per  hour  =  22  feet  per  second. 

Thus,  using  foot-second  units,  we  have 

v^  =  m2  +  2  gJi  =  222  +  2-32-60  =  4324, 
so  that  i;  =  66  feet  per  second,  approximately, 

=  45  miles  per  hour. 

Thus  the  velocity  of  the  bicycle,  if  unchecked  by  friction  or  air  resist- 
ance, would  be  one  of  about  45  miles  per  hour. 


EXAMPLES 

1.  An  automobile  running  40  miles  an  hour  comes  to  the  foot  of  a  steep  hill, 
and  at  the  same  instant  the  engine  is  shut  off.  To  what  height  up  the  hill  will 
the  automobile  go  before  coming  to  rest  (neglect  friction,  etc.)? 

2.  A  laborer  has  to  send  bricks  to  a  bricklayer  at  a  height  of  10  feet.  He 
throws  them  up  so  that  they  reach  the  bricklayer  with  a  velocity  of  10  feet  per 
second.  What  proportion  of  his  work  could  he  save  if  he  threw  them  so  as  only 
just  to  reach  the  bricklayer  ? 

3.  A  gun  carriage  of  mass  3  tons  recoils  on  a  horizontal  plane  with  a  velocity 
of  10  feet  per  second.  Find  the  steady  pressure  that  must  be  applied  to  it  to 
reduce  it  to  rest  in  a  distance  of  3  feet. 

4.  A  ship  of  2000  tons  moving  at  30  feet  a  minute  is  brought  to  rest  by  a 
hawser  in  a  distance  of  2  feet.    Find  in  tons  what  pull  the  hawser  has  to  sustain. 


174  AVORK 

5.  A  bicycle  and  rider  weigh  200  pounds,  and,  -wdien  riding  along  a  level 
road  at  25  miles  an  hour,  the  rider  suddenly  applies  a  brake  which  presses 
on  the  tire  with  a  foi'ce  equal  to  the  weight  of  50  pounds.  If  tlie  coefficient  of 
friction  between  the  brake  and  the  tire  is  J,  find  how  far  the  machine  will  go 
before  coming  to  rest. 

6.  In  the  last  question,  how  far  will  the  machine  go  if,  instead  of  the  road 
being  level,  it  is  down  an  incline  of  1  in  20? 

7.  A  bullet  fired  with  a  velocity  of  1000  feet  per  second  penetrates  a  block 
of  wood  to  a  depth  of  twelve  inches.  Prove  that  if  it  were  fired  through  a  board 
of  the  same  wood,  two  inches  thick,  its  velocity  on  emergence  would  be  about 
913  feet  per  second.  (Assume  the  resistance  of  the  wood  to  the  bullet  to  be 
constant.) 

8.  Two  equal  weights  P  and  P  are  supported  by  a  string  passing  over  two 

small  smooth  pulleys  A  and  B  in  the  same  horizontal  line,  and  a  weight 

2 
W  =  —  P  is  attached  to  the  middle  point  of  the  string  between  A  and  B. 

.   Vi 

Prove  that  W  will  continue  to  descen4  until  WAB  forms  an  equilateral  triangle, 
and  examine  what  will  happen  after  this. 

9.  A  string  of  natural  length  I  and  modulus  X  is  suspended  between  two 
points  A,  B  in  the  same  horizontal  line  and  at  a  distance  h  apart,  and  lias  a 
weight  W  attached  to  its  middle  point.  The  weight  W  is  held  at  rest  midway 
between  the  points  A,  B,  and  is  suddenly  set  free.  Find  how  far  it  will  fall 
before  being  brought  to  rest  by  the  strings. 

10.  A  heavy  particle  hangs  by  a  string  of  natural  length  I,  which  it  stretches 
to  a  length  I',  the  other  end  of  the  string  being  fixed.  The  particle  is  pulled 
down  to  a  length  2  V  below  the  point  of  support,  and  is  then  set  free.  How  high 
will  it  rise  ? 

11.  Determine  the  horse  hower  which  could  be  obtained  from  the  kinetic 
energy  of  a  river  at  a  place  where  the  width  is  100  feet,  the  mean  depth  20  feet, 
and  the  mean  velocity  4^  miles  per  hour.  (A  cubic  foot  of  water  weighs  62.5 
jjounds.) 

12.  The  river  of  the  last  question  ends  in  a  waterfall  of  which  the  bottom 
is  50  feet  below  the  river  bed.  Find  the  horse  power  which  could  be  obtained 
from  the  water. 

13.  A  locomotive  burns  li-  pounds  of  coal  per  hor.se-power-hour.  How  much 
coal  must  be  burned,  beyond  that  consumed  in  overcoming  gravity,  friction,  etc., 
in  giving  to  a  train  of  300  tons  a  velocity  of  55  miles  an  hour  ? 

Stable  and  Unstable  Equilibrium 

147.  Let  us  consider  a  system  at  rest  in  a  position  of  equilibrium, 
and  capable  of  moving  from  this  position  by  only  one  path,  over 
which  it  may,  of  course,  move  in  either  direction.  As  an  illustra- 
tion of  a  system  of  this  kind  we  may  take  a  locomotive  standing 


STABILITY  AND  INSTABILITY  175 

on  a  pair  of  rails,  a  door  turning  about  a  hinge,  or  a  bead  sliding 
on  a  wire.  The  system  is  supposed  to  be  acted  on  l)y  any  number 
of  conservative  forces,  but  to  be  in  a  position  of  eciuililirium  under 
these  forces. 

Let  P  denote  the  position  of  equilibrium,  and  let  //;,  l)e  the 
potential  energy  when  the  system  is  in  configuration  P.  Let  x 
denote  any  coordinate  which  measures  how  far  the  configuration 
of  the  system  has  moved  from  P  —  for  instance,  returning  to  our 
former  illustrations,  x  might  denote  the  distance  the  locomotive  had 
moved  along  the  track,  the  angle  through  wliicli  the  door  had 
turned  about  its  liinges,  or  the  distance  the  bead  had  moved  along 
the  wire.  The  value  of  x  will  of  course  be  considered  positive  if  the 
system  moves  in  one  direction,  and  negative  if  it  moves  in  the  other. 

As  the  system  moves  away  from  its  equilibrium  configuration 
P,  the  value  of  x  will  change.  The  value  of  W,  the  potential 
energy,  will  also  change,  and  as  it  depends  only  on  the  value  of  x  if 
the  forces  are  conservative,  we  may  say  that  W  is  a  function  of  x. 

By  a  well-known  theorem,  we  can  expand  W  in  powers  of  x  in 
the  form  /  >     \ 

in  which  the  subscript  P  denotes  (as  it  has  already  been  supjiosed 
to  denote  in  the  case  of  Tr^)  that  the  quantity  is  to  be  evaluated 
in  the  configuration  P.  Since  the  configuration  P  is  sui)posed  to 
be  one  of  equilibrium,  we  have  by  the  theorem  of  §  135, 

CX  Jp 

so  that  equation  (40)  becomes 

'r=,r-  +  i.=  (|^)^+....  (41) 

For  configurations  near  to  P,  x  is  small,  so  that  the  term 
\^\ )  in  equation   (41),  although    itself    small,  is  vet  verv 

\CJ?  jp 

large  compared  with  the  terms   in  x-',  x^,  etc.,  which  follow  it. 


17G  WORK 

Thus,  for  configurations  near  to  P,  we  may  neglect  these  latter 
terms  altogether,  and  write  the  equation  in  the  form 

W-^^]>  =  \^i^-  (42) 


The  value  of  (  — -  J  may  be  either  positive  or  negative. 

If  it  is  positive,  then  TV—  IJ],  is  positive  whatever  the  value  of 
X,  so  that  the  potential  energy  W  in  every  configuration  near  to 
P  is  greater  than  that  in  configuration  P.  In  other  words,  IV  is  a 
Tniniiniiin  at  P. 

So  also  if  ( ^—-  \  is  negative,  W  —  Wp  is  negative  for  all  small 


/p 
values  of  x,  and  we  find  that  W  is  a  maximum  at  P. 

148.  Suppose  now  that  the  system  is  placed  at  rest  in  some  con- 
figuration near  to  P.  This  configuration  is  not  one  of  equilibrium, 
so  that  the  system  cannot  remain  at  rest.  To  determine  the  direc- 
tion in  which  it  begins  to  move,  we  need  only  notice  that  as  the 
system  moves  it  acquires  kinetic  energy,  and  as  this  must,  by 
§  143,  be  acquired  at  the  expense  of  its  potential  energy,  we  see 
that  the  system  will  begin  to  move  in  such  a  direction  that  its 
potential  energy  will  be  diminished. 

A  glance  at  equation  (42)  will  show  whether  this  direction  is 

towards  or  away  from  P.    We  see  that  if  ( -—;■ )  is  positive,  the 

\  ^^^"  Ip 
value  of  x"^  must  decrease,  so  that  the  motion  will  be  towards  P, 

whatever  the  value   of   x.     Similarly,  if   ( ^^-^  |  is   negative,  the 

value  of  x^  must  increase,  so  that  the  motion  will  be  always  away 
from  P. 

We  have  now  seen  that  if  the  system  is  placed  in  a  con- 
figuration adjacent  to  P,  the  question  of  whether  the  motion 
which  ensues  is  towards  or  away  from  P  does  not  depend  on  the 
configuration  in  which  the  system  is  placed,  but  depends  on  the 

sign  ot  ( -— 


STABILITY  AND  INSTABILITY  177 

We  have  seen  that  if  /'  is  a  configui-ation  of  equilibrium,  and 
if  the  system  is  slightly  displaced  from  P  to  a  ueighboring  con- 
figuration, then 

(a)  if  I  -— -  j  IS  positive,  the  system,  when  set  free,  will  return 
to  its  original  position  of  equilibrium ; 

(h)  if  (  — ^ )  is  negative,  the  system  when  set  free  will  move 
farther  away  from  its  original  position  of  equilibrium. 

Equilibrium  of  the  first  kind  is  called  stable  equilibrium ;  equi- 
librium of  the  second  kind  is  called  tmstahle  equilibrium. 
We  can  summarize  the  results  as  follows : 


\  ca;-  Jp 

Potential  Energy  W 

Equilibrium 

+ 

minimum 
maximum 

stable 
unstable 

149.  Theorem.  Positions  of  stable  and  unstable  equilibriuTn 
occur  alternately. 

We  can  assume  that  we  are  dealing  only  with  finite  forces,  so 
that  the  function  W  will  always  be  finite :  it  can  never  pass 
through  the  values  Tr=±co.  It  must  be  continuous,  for,  by 
hypothesis,  the  work  done  in  placing  the* S3-stem  in  any  configura- 
tion must  have  a  definite  value,  so  that  the  potential  energy  can 
have  only  one  value  for  a  given  configuration.  Also  the  differential 
coefficients  of  the  potential  energy  must  be  finite,  for  these  measure 
the  forces  (§  133)  which  can  have  only  finite  values  in  any  given 
configuration. 

Thus  if  the  graph  of  the  function  W  is  drawn,  we  see  that  it 
must  consist  of  portions  in  which  W  is  alternately  mcreasing  and 
decreasing.  On  passing  from  a  portion  in  which  If  increases  to 
one  in  which  it  decreases,  we  pass  through  a  point  at  which  W  is 
a  maximum,  wliile  in  passing  from  a  region  in  wliich  W  decreases 


178 


WOKK 


to  one  in  which  it  increases,  we  pass  through  a  minimum.  Tims 
maximum  and  minimum  vahies  of  W  must  occur  alternately,  or, 
what  is  the  same  thing,  configurations  of  stable  and  unstable 
equilibrium  must  occur  alternately. 

150.  Examples  of  these  two  kinds  of  equilibrium  can  be  found 
in  tlie  illustrations  already  employed. 

1.  Locomotive  moving  on  a  pair  of  rails.  Let  h  be  the  height  of  the  center 
of  gravity  in  any  position,  let  x  denote  distances  measured  horizontally 
along  the  track,  and  let  M  be  the  mass  of  the  locomotive.    The  potential 

energy  is  then  Mg/i.  The  con- 
dition for  equilibrium  in  the 
configuration  x  —  0  is 


-W.)=0, 


(Ih 


<lx 


h  maximum 
Equil. unstable 


h  minimum 
Equil.  stable 


h=0 


Fig.  97 


0,   expressing'    that    the 

value  of  h  must  be  either  a  maxi- 
mum or  a  minimum.  The  table 
on  page  177  shows  that  if  h  is  a 
minimum, — -i.e.  if  the  center  of 
gravity  is  at  its  lowest  point,  — 
the  equilibrium  will  be  stable. 
Thus,  if  the  locomotive  is  moved  slightly  from  this  position,  it  will  roll 
back  to  it  again.  If  h  is  a  maximum,  —  i.e.  if  the  center  of  gravity  is  at 
its  highest  point,  —  the  equilibrium  will  be  unstable.  The  locomotive  is 
now  at  the  summit  of  a  hill,  and  if  displaced  to  either  side  of  the  summit, 
will  continue  rolling  down  the  hill. 

Note.  If  the  moving  parts  of  the  engine  are  not  "  balanced  "  properly,  the  center 
of  gravity  may  not  always  be  at  the  same  height  above  the  rails,  so  tliat  the  maxima 
and  minima  of  h  do  not  necessarily  occur  at  points  wliere  tlie  heiglit  of  the  track  is  a 
maximum  or  a  minimum.  For  instance,  a  position  of  equilibrium  might  occur  where 
the  track  was  not  level,  or  again  a  position  of  stable  equilibrium  miglit  occur  at  a 
point  at  which  the  track  was  at  its  highest  point,  the  height  of  the  center  of  gravity 
above  the  rails  being  of  course  a  minimum  at  this  point.  Thus  if  the  engine  were  dis- 
placed to  a  point  sliglitly  lower  on  the  track,  and  set  free,  it  would  return  of  itself  to 
the  highest  point.  The  principle  here  is  the  same  as  that  of  mechanical  toys  whicli, 
on  being  placed  at  rest  at  the  foot  of  an  inclined  plane,  start  to  roll  up  the  plane  as 
soon  as  set  free. 

We  notice  that  positions  of  stabie  and  unstable  equilibrium  must  occur 
alternately,  as  already  proved  in  §  149. 

2.  Door  turning  on  hinges.  Here  again  the  potential  energy  is  High,  where 
Ji  is  the  height  of  the  center  of  gravity  of  the  door  above  any  standard 


STABILITY  AND   INSTABILITY 


179 


level.  As  the  door  turns  on  its  hinges,  its  center  of  gravity  describes  a 
circle  about  the  line  of  hinges.  If  this  line  is  perfectly  vertical,  the  circle 
described  by  the  center  of  gravity 
lies  entirely  in  a  horizontal  plane,  so 
that  every  position  is  one  of  equilib- 
rium, and  the  question  of  stability  or 
instability  does  not  arise.  If,  how- 
ever, the  line  of  hinges  is  not  per- 
fectly vertical,  the  circle  will  lie  in 
an  inclined  plane.  The  points  at 
which  the  height  above  the  standard 
horizontal  plane  is  a  maximum  or 
minimum  are  two  in  number  : 

P,  the  highest  point  of  the  circle.  Fig.  98 

at  which  equilibrium  is  unstable  ; 

Q,  the  lowest  point  of  the  circle,  at  which  equilibrium  is  stable. 

3.  Bead  sliding  on  wire.  To  obtain  a  definite  problem,  let  us  svipjjose 
that  the  bead  P  slides  on  an  elliptic  wire  placed  so  that  its  major  axis  A  A' 
is  vertical,  and  let  it  be  acted  on  by  its  weight,  and 
also  by  the  tension  of  a  stretched  elastic  string  of 
which  the  other  end  is  tied  to  the  center  of  the 
ellipse.  Let  a,  6  be  the  semi-axes  of  the  ellipse,  and 
let  I,  X  be  the  natural  length  and  modulus  of  the 
string,  /  being  greater  than  «,  so  that  the  string  is 
always  stretched.  Let  tv  be  the  w^eight  of  the  bead. 
The  first  step  is  to  calculate  the  potential  energy 
in  any  configuration.  Let  the  configuration  be 
sjiecified  by  the  eccentric  angle  <^  of  the  point  on 
the  ellipse  occupied  by  the  bead.  The  height  of 
the  bead  above  the  center  of  the  ellipse  is  then 
acos^,  so  that  that  part  of  the  potential  energy 
which  arises  from  gravitational  forces  is  wa  cos  ^. 
The  length  of  the  string  r  is  given  by 

7-2  =  rt2  cos2  <l>  +  h'  sin2  0,  (a) 

and  the  work  done  in  stretching  the  string  from  leug-th  I  to  length  r  is  (§  1 13) 

ir-. 


ho- 


This  may  be  taken  to  be  the  part  of  the  potential  energy  whioli  arises 
from  the  stretching  of  the  string.      Tlius  the  total  potential  energy  will  be 


TT' 


<p  +  ^fO--o-- 


(&) 


180 


WOKK 


The  positions  of  equilibrium  are  now  given  by  ——  =  0,  or 

d<f> 

ion  sin  ^ (r  —  I)  —  =0, 

/  d(t> 

or,  substituting  for  r  from  equation  (a), 

wa  sin0  +  -(o^  —  b-)  sin^  cos0 ^  '  -  =  0. 

!•  va2  cos^^  +  i'^sin"'*^ 

Rationalizing,  we  find  that  roots  are  given  by  sin^  =  0,  and  also  by 
>ca  +  -  ((fi  —  b^)  cos (p     ((fl  cos^  <f>  -]-  h^  sin-  (^)  —  X-  (o^  _  Jj'^y  cos^  0  =  0, 

which  reduces  to 

[ira  +  -  («2  -  b-)  cos  0]  ""  r(rt2  -  ^2)  C0S2  0  +  //^l  -  X2  (a2  _  i2-)2  cos2  0  =  0,     (c) 

an  equation  of  the  fourtla  degree  in  cos  0. 

The  roots  of   sin  0  =  0  are  0  =  0,  tt,  so    that   there    are   always   two 

positions  of  equilibrium  at  A,  A',  the  ends  of  the  major  axis.    Equation 

(c),  being  of  the  fourth  degi-ee,  nuiy  have 
0,  2,  or  4  real  roots  in  cos  0.  The  equa- 
tion as  it  stands  has  been  obtained  by 
squaring  both  sides  of  the  equation  to 
be  satisfied,  and  in  doing  this  we  have 
doubled  the  number  of  roots  of  the  true 
equation.  Thus  the  true  equation  will 
only  be  satisfied  by  0,  1,  or  2  real  roots 
in  cos  0.  In  other  words,  between  A  and 
A',  on  either  side  of  the  wire,  there  can 
be  at  most  two  i:)ositions  of  equilibrium. 
It  would  be  a  tedious  piece  of  work  to  find  the  actual  values  of  the  roots 

d'^W 
for  cos  0,  and  then  determine  the  signs  of  the  values  of correspond- 

^/02 

ing  to  these  roots.  The  question  is,  however,  very  much  simplified  by 
using  the  general  theory  of  stable  and  unstable  configurations. 

If  we  put  X  =  0  in  expression  (J),  we  obtain  as  the  jiotential  energy  in 
the  case  in  which  X  is  A'anishingly  small  in  comparison  with  id, 

W  =  uja  COS0, 

of  which  the  graph  is  shown  in  fig.  100.  Here  there  are  only  two  positions 
of  equilil)rium,  namely  0  =  0  and  0  =  tt,  the  former  being  unstable  (f/) 
and  the  latter  stable  (.V). 


STABILITY  AXD   INSTABILITY 


181 


A 

.     / 

^ 

\  / 

u 

Y 

u 

V 

1 

s 

— \ — 

1 

f>=0 


0= 

Fig.  101 


(p—'2Tr 


Again,   if  we  jiut  w  =  0  in  expression   (//),  we  obtain  as  the  potential 
energy  in  the  case  in  which  X  is  infinitely  great  in  comparison  with  w, 

and  the  graph  of  W  in  this  case  is  sliown  in  fig.  101.    There  are  four  posi- 
tions of  equilibrium,  n      tt  '^' "" 

0  =  0,       -,       IT,       —^, 

which  are  respectively  unstable,  stable,  unstable,  and  stable. 

The  general  case  in  which  \  stands  in  a  finite  ratio  to  w  is  intermediate 
between  the  two  extreme  cases  which  have  been  considered.  The  graph 
for  W  in  the  general  case  can  be  obtained  by  compounding  the  two  graphs 
already  drawn.  To  obtain  the  ordinate  corresponding  to  any  value  of  <(>, 
we  multiply  the  corresponding  ordinates 
in  the  graphs  already  obtained  by  the 
appropriate  constants,  and  add.  The  two 
ordinates  give  the  two  terms  of  expres- 
sion (b)  separately  :  their  sum  gives  the 
total  value  of  W  as  required. 

From  this  geometrical  construction  it 
is  clear  that  0  =  0  remains  a  configura- 
tion of  unstable  equilibrium.  The  con- 
figuration (/>  =  TT  is  also  a  configuration  of 
equilibrium,  but  may  be  either  stable  or  unstable.  Between  these  two 
configurations  there  may  be  one  other  configuration  of  equilibrium,  as  in 
fig.  101  ;  or  there  may  be  none,  as  in  fig.  100.  Since,  by  §,119,  stable  and 
unstable  configurations  occur  alternately,  it  is  clear  that  if  the  configuration 
^  =  TT  is  stable,  there  can  be  no  other  configuration  of  equilibrium  between 
this  and  0  =  0,  while  if  0  =  7r  is  unstable,  there  must  be  one  configuration 
of  equilibrium  between  0  =  tt  and  <f>  =  0,  and  this  must  be  stable. 

The  stability  or  instability  of  the  configuration  (/>  =  tt  accoi-dingly  deter- 
mines the  nature  of  the  solution  for  a  given  value  of  X.    This  stability  or 

instability  is  in  turn  determined  bv  the  sign  of at  </>  =  tt.     To  deter- 

mine  this,  let  us  write  tr  —  <f>  =  d  near  to  0  =  tf,  and  neglect  terms  smaller 
than  6^.    We  have,  to  this  approximation, 

r^  =  (fi  cos-<p  +  Ifi  sin^^ 
=  (fl  -  (r/2  -  /;2)  sin2^ 
=  «•-  -{(fl  -  lr)e-, 
so  that  by  equation  (!>'), 

W  =  ica  cos  <p  -^  ^{r  -  If 

=-,™(.-;i.A[„(,-i'i!^%=)-,j. 


IS'2  WOKK 


Thus  _-  =  ..a--^ -f 1. 


It  follows  that  equilibrium  at  ^  =  tt  is  stable  or  unstable  according  as 

tva-l 

X  <  or  > 

(a-^  -  b-^){a  -  I) 

To  sum  up,  there  are  two  cases  : 

I.  X  < The   only  positions  of  equilibrium   are  ^  =  0 

(a^  —  o'^)(a  — /) 

and  (p  =  TT,  which  are  respectively  unstable  and  stable. 

II.  X  > There  are  positions  of  equilibrium  0  =  0  and 

(a^  _  ^^^(a  -  /) 

0  =  TT,  both  unstable,  and  also  an  intermediate  position  which  is  stable. 
This  last  position  is  determined  by  equation  (c). 


Critical  and  Neutral  Uquilihrium 

f]1  TIT' 

151.  If  the  value  of  -^^-j^  at  a  position  of  equilibrium  is  zero, 

the  equilibrium  is  called  critical.  So  far,  we  have  not  discovered 
what  happens  when  a  system  is  slightly  displaced  from  a  posi- 
tion of  critical  equilibrium. 

In  general,  the  value  of  W  in  the  neighborhood  of  any  position 
of  equilibrium  can  be  expanded  in  the  form  (cf.  equation  (41)) 

--..l.<f|}H-|.(|^).A.(|^)^.....      (43) 

If  -—J  vanishes  at  P,  the  most  important  term  in  the  value  of 
W  —  Wp  is  that  in  5c',  so  that  we  have  approximately 

Here  W  —  Wp  changes  sign  on  passing  through  a;  =  0,  the  con- 
figuration of  equilibrium,  so  that  the  graph  of  W  is  as  shown  in 


STABILITY  AND  INSTABILITY  183 

fig.  102,  having  a  horizontal  tangent  and  point  of  inflection  at  P. 
On  one  side  the  potential  energy  is  less  than  at  P,  on  the  other 
side  it  is  greater. 

Let  Q,  Q'  be  two  adjacent  configurations  on  these  two  sides 
of  P.    If  the  system  is  placed  at  Q,  it  must  move  so  that  its  poten- 
tial energy  decreases,  and  therefore  moves  away  from  P.    If  it  is 
placed  at  Q' ,  for  the  same  reason  it  must  move 
at  first  towards  P,  but  it  will  move  beyond  P 
and  will  then  continue  to  move  away  from  P, 
—  for  it  cannot  come  to  rest  until  its  potential 
energy  is   again   equal  to  that   at  Q' ,  and  this  " 

cannot  happen  m  the  neighborhood  of  P.    Thus 
if  the  system  starts  from  any  configuration  in  the  neighborhood 
of  P,  it  will  ultimately  be  movmg  away  from  P.    In  other  words, 
the  equilibrium  is  unstable. 

d-JV 
Thus  if  — -  =  0  at  P,  the  equilibrium  is,  in  general,  unstable. 

An  exception  has  to  be  made  when  -— -  =  0;  for  then  we  have 


This  case  may  be  treated  as  in  §  148,  and  we  find  that  the 

^r^j  is  positive 

Ui   iici^auivc.  "     /P 

152.  Higher  degrees  of  singularity  may  be  treated  in  the  same 
way,  and  we  easily  obtain  the  following  general  rules : 

If  the  first  differential  coeffiGient  which  does  not  vanish  is  of  odd 
order,  the  equilibrium  is  unstable. 

If  the  first  differential  coeffcient  ivhich  does  not  vanish  is  of  even 
order,  the  eq^iilibrium  is  stable  or  unstable  according  as  this  differ- 
ential coeffcient  is  positive  or  negative. 

It  is  possible  for  all  the  differential  coefificients  to  vanish,  in 
which  case  the  problem  is  best  treated  by  other  methods. 


184  WORK 

For  instance,  if  the  potential  energy  is  of  the  form 

W  =  x'e  ■<% 

it  will  be  found  that  all  the  differential  coefficients  vanish  in  the 
configuration  given  by  «  =  0.  On  drawing  a  graph  of  the  function 
W  it  appears  that  the  equilibrium  is  stable. 

It  may  be  that  all  the  differential  coefficients  of  W  vanish 
because  W  is  a  constant  throughout  the  whole  of  a  range  sur- 
rounding the  configuration  under  consideration.  If  this  is  so, 
the  system  may  be  displaced,  and  there  will  be  no  force  tending 
to  move  it  from  its  new  configuration  —  every  configuration  is 
one  of  equilibrium.  Equilibrium  of  tliis  kiud  is  called  neutral 
equilibrium. 

A  case  of  neutral  equilibrium  has  already  occurred  in  Ex.  2,  p.  179,  —  a 
door  free  to  swing  about  a  vertical  line  of  hinges.  A  second  case  is  that 
of  a  sphere  rolling  on  a  horizontal  plane. 

Systems  possessing  Several  Degrees  of  Freedom 

153.  So  far  we  have  considered  only  systems  wliich  are  limited 
to  moving  through  a  smgle  series  of  configurations  —  systems 
with  only  a  single  degree  of  freedom.  The  determination  of  the 
stability  or  instabihty  of  a  system  having  more  than  one  degree 
of  freedom  is  a  more  complex  problem. 

If  the  potential  energy  is  absolutely  a  minimum  in  a  position 
of  equilibrium,  so  that  every  possible  motion  involves  an  increase 
of  potential  energy,  then  the  equilibrium  is  stable.  This  obviously 
can  be  proved  by  the  same  argument  as  has  served  when  there  is 
only  one  degree  of  freedom. 

If  the  potential  energy  is  not  an  absolute  minimum,  —  that  is 
to  say,  if  displacements  are  possible  in  which  the  potential  energy 
decreases  wliile  moving  away  from  the  position  of  equilibrium, — 
then  the  coutiguration  is  one  of  unstable  equilibrium.  This  will  be 
proved  later.  It  cannot  be  proved  by  the  methods  used  m  this  chap- 
ter, and  so  we  defer  the  question  until  later  (Chapter  XII). 


EXAMPLES  185 

GENERAL  EXAMPLES 

1.  Prove  that  the  horse  power  of  an  engine  which  overcomes  a  resist- 
ance of  R  pounds  at  a  speed  of  .S^  miles  an  hour  is 

RS  ^  375. 

2.  A  train  weighing,  with  the  locomotive,  500  tons  is  kept  moving  at 
the  uniform  rate  of  30  miles  an  hour  on  the  level,  the  resistance  of  air, 
friction,  etc.,  being  40  pounds  per  ton.    Find  the  horse  power  of  the  engine. 

By  how  much  must  this  horse  power  be  increased  if  the  rate  is  to  be 
maintained  while  water  is  taken  up  from  a  trough  between  the  rails  to  the 
amount  of  20  pounds  per  foot  passed  over,  the  height  to  which  the  water  is 
raised  above  the  trough  being  10  feet,  and  the  kinetic  energy  imparted  to 
the  water  in  the  trough,  as  well  as  that  of  the  motion  of  the  water  taken  up, 
relatively  to  the  tank,  being  neglected  ? 

3.  The  sides  of  a  conical  hill  are  of  such  a  shape  that  a  given  mass  will 
just  rest  on  them  without  slipping.  A  man  wishes  to  move  this  mass  from 
a  point  at  the  base  of  the  hill  to  a  second  point  diametrically  opposite  to 
the  first.  Show  that  the  work  of  dragging  it  over  the  hill  is  less  than  the 
work  of  dragging  it  round  the  base  of  the  hill,  in  the  ratio  2  :  tt. 

4.  Show  that  the  work  a  man  does  in  dragging  a  weight  up  a  hill  from 
a  given  point  A  to  the  summit  B  depends  only  on  the  positions  of  A  and 
B,  and  is  independent  of  the  shape  of  the  hill,  provided  he  keeps  always 
in  the  vertical  plane  through  A  and  B. 

5.  A  catapult  is  made  by  tying  the  two  ends  of  a  piece  of  elastic,  natural 
leng-th  «,  modulus  of  elasticity  X,  to  the  two  prongs  of  a  forked  piece  of 
wood,  distant  I  apart,  I  being  greater  than  a.  A  stone  of  mass  m  is  placed 
at  the  middle  point  of  the  catapult,  and  is  drawn  back  until  the  string  is 
stretched  to  double  its  natural  length.  If  it  is  then  set  free,  find  the 
velocity  with  which  it  will  leave  the  catapult. 

6.  If,  in  tlie  last  question,  the  stone  is  projected  vertically  upwards  from 
the  catapult,  find  the  height  to  which  it  will  rise  before  coming  to  rest. 

7.  A  necklace  of  mass  m  is  made  of  beads  threaded  on  a  light  string  of 
modulus  X.  It  is  held  in  a  horizontal  plane,  with  the  string  unstretched, 
resting  on  the  surface  of  a  smooth  right  circular  cone  of  semivertical  angle 
a,  of  which  the  axis  is  vertical.  If  the  necklace  is  let  go,  how  far  will  it 
slip  down  the  cone  before  coming  to  rest  ? 

8.  A  fly  wheel  is  of  radius  2  feet  6  inches,  and  the  weight  of  the  spokes, 
etc.,  may  be  neglected  in  conijiarison  with  that  of  the  rim.  It  is  rotating 
at  the  rate  of  250  revolutions  per  minute  about  a  fixed  axle,  which  is 
3  inches  in  diameter,  the  coefficient  of  friction  between  the  wheel  and  axle 
being  ^l.  If  it  is  left  to  itself,  find  how^  many  revolutions  it  w^ill  make 
before  stopping. 


V{^ 


186  ■  WORK 

9.  A  sjiider  hangs  from  the  ceiling  by  a  thread  of  moduhis  of  elasticity 
equal  to  its  weight.  Show  that  it  can  climb  to  the  ceiling  with  an  expend- 
iture of  work  equal  to  only  tlu-ee  quartei's  of  what  would  be  retiuired  if 
the  thread  were  inelastic. 

10.  A  fine  thread  having  two  masses  each  equal  to  P  suspended  at  its 
ends  is  hung  over  two  smooth  pegs  in  the  same  horizontal  line,  distant 
2  a  apart.  A  mass  Q  is  then  attached  to  the  middle  point  of  the  portion  of 
the  string  between  the  pegs  and  allowed  to  descend  imder  gravity.  Show 
that  its  velocity  after  falling  a  depth  x  will  be 

2g(x^  +  a?)  (Qx  +  2  P«  -  2  P  Vx'^  +  a^)  ] 
Q(x'^  +  a')  +  2  Pjf'  j 

11.  Assuming  that  the  attraction  of  the  earth  on  a  body  outside  the 
earth  falls  off  inversely  as  the  square  of  the  distance  of  the  body  from  the 
earth's  center,  find  with  what  velocity  a  shot  would  have  to  be  fired  ver- 
tically upwards  from  the  earth's  surface  so  as  never  to  return  to  the  earth 
at  all. 

12.  A  steam  hammer  of  weight  30  tons  is  pressed  down  partly  by  its 
weight  and  partly  by  the  pressure  of  steam  in  a  vertical  cylinder  acting  on 
a  piston  which  moves  with  the  hammer.  The  area  of  the  piston  is  4  square 
feet,  and  the  steam  pressure  is  22.3  pounds  to  the  inch.  If  the  hammer  is 
raised  a  height  of  2  feet  above  its  block,  and  set  free,  find  the  velocity  with 
which  it  will  strike  the  block. 

13.  The  ends  of  a  uniform  rod  of  length  I  are  connected  by  a  string  of 
length  a  which  is  jilaced  over  a  smooth  peg.  Show  that  the  rod  can  only 
hang  in  a  horizontal  or  vertical  position,  and  examine  the  stability  or 
instability  of  these  positions. 

14.  Two  equal  uniform  rods  are  rigidly  jointed  in  the  shape  of  the  let- 
ter L,  and  placed  astride  a  smooth  circular  cylinder  of  radius  a.  Find 
the  smallest  length  of  the  rods  consistent  with  stability  of  e(][uilibrium,  the 
rods  being  constrained  to  remain  in  a  vertical  plane  perpendicular  to  the 
axis  of  the  cylinder. 

15.  A  cube  of  stone  of  edge  a  rests  symmetrically  and  with  its  base 
horizontal  on  a  rough  circular  log  of  diameter  b.  Show  that  the  equilib- 
rium is  stable  or  unstable  according  as  ft  >  or  <a. 

16.  A  rocking  stone  rests  on  a  fixed  stone,  the  contact  being  rough,  and 
the  common  normal  at  the  point  of  contact  being  vertical.  If  p,  p'  be 
the  radii  of  curvature  of  the  surfaces  of  the  two  stones  at  the  point  of 
contact,  and  if  h  be  the  height  of  the  center  of  gravity  of  the  movable  stone, 
show  that  the  equilibrium  of  the  rocking  stone  will  be  stable  or  unstable 

according  as  111 

-  >  or  <  -  -I-  -• 
/*  p      p 


EXAMPLES  187 

17.  A  ladder  of  length  h  and  weight  w  stands  in  a  vertical  i^osition  on 
a  rough  floor,  an  elastic  string  being  tied  to  its  topmost  point  and  to  a 
point  in  the  ceiling  at  a  height  h  above  the  floor,  its  tension  being  T.  Show 
that  the  equilibrium  is  stable  or  unstable  according  as 

T  >  or  <  -^ — . ^  • 

•21, 

18.  If  the  tension  in  questicm  17  is  equal  to  w  (h  —  h)/2h,  determine 
whether  the  equilibrium  is  stable  or  unstable. 

19.  If  in  question  14  the  rods  are  of  the  critical  length  which  separates 
stability  from  instability,  show  that  the  equilibrium  is  neutral,  so  that  the 
rods  can,  within  certain  limits,  rest  in  any  position  in  the  plane  perijeu- 
dicular  to  the  axis  of  the  cylinder. 

20.  Show  that  the  rods  in  the  last  question  are  in  stable  equilibrium 
as  regards  displacements  in  which  the  plane  of  the  rods  rotates  about  a 
vertical  axis,  and  find  the  couple  required  to  hold  the  rods  in  a  position  in 
which  the  plane  makes  any  given  angle  d  with  the  axis  of  the  cylinder. 

21.  Find  the  smallest  length  of  the  rods  in  the  last  question,  in  order 
that  the  equilibrimn  nuiy  be  stable  for  all  possible  displacements. 

22.  The  radii  of  curvature  at  the  blunt  and  jiointed  ends  of  a  hard-boiled 
egg  are  a  and  b  respectively,  and  the  egg  can  just  be  made  to  balance  on 
its  blunt  end  when  stood  on  a  rough  horizontal  surface.  Show  that  it  can 
be  made  to  balance  on  its  pointed  end  if  stood  inside  a  hemispherical  basin 
of  radius  less  than  .        ,  ^ 

a  -\-  b  —  c 

where  c  is  the  longest  axis  of  the  egg.     If  the  radius  of  the  basin  is  ju.st 
equal  to  «(e  —  h)/{fi  +  h  —  r),  would  the  equilibrium  be  stable  or  unstable  ? 

23.  ^4,  B,  Care  three  equidistant  smooth  pegs  in  the  same  horizontal 
line,  and  a  heavy  uniform  string  has  its  ends  tied  to  A,  C,  and  is  looped 
over  B.  Show  that  there  may  or  may  not  be  a  position  of  equilibrium  in 
which  the  two  catenaries  AB,  BC  are  xmequal,  and  that  if  there  is  such  a 
position  it  will  be  stable. 

Show  also  that  the  position  of  equilibrium  in  which  the  middle  point  of 
the  string  is  at  B  is  unstable  or  stable  according  as  an  uusymmetrical 
position  of  equilibrium  does  or  does  not  exist. 


CHAPTEE  VIII 
MOTION  OF  A  PARTICLE   UNDER  CONSTANT  FORCES 

154.  The  simplest  case  of  motion  of  a  single  particle  occurs 
when  the  particle  is  acted  upon  only  by  constant  forces  and  moves 
in  a  straight  line. 

If  P  is  the  component  force  in  the  direction  of  the  motion  of 

the   particle,  there  will,  by  the    second    law  of  motion,  be  an 

acceleration  /  given  by 

P  =  mf, 

where  m  is  the  mass  of  the  particle.    Since  the  forces  are,  by 
hypothesis,  constant,  the  acceleration  /  is  also  constant. 

Let  the  particle  start  with  a  velocity  it,  and  move  with  a  con- 
stant acceleration  /.  In  time  t  the  increase  in  velocity  is  ft,  so 
that,  after  any  time  t,  the  whole  velocity  is  %  +  ft.    Denoting  tliis 

velocity  by  v,  we  have 

v  =  u+ft.  (44) 

By  definition,  v  is  equal  to  -j-^  where  s  is  the  space  described 
from  the  beginning  of  the  motion.    We  accordingly  have 

an  equation  giving  the  rate  of  increase  of  s  at  any  instant.    Integrat- 
ing, we  obtain 

s  =  tu  +  \ft\  (45) 

no  constant  of    integration  being  needed,  because   the    distance 
described  at  time  ^  =  0  has  to  be  0,  from  the  definition  of  s. 

By  equation  (44),  n  —  v  —ft,  so  that  equation  (45)  can  be  written 

s  =  vt-  \ft\  (46) 

188 


MOTION  UNDER  CONSTANT  FORCES  189 

This  gives  the  distance  described  in  time  t,  when  we  know  the 
velocity  v  with  which  the  particle  arrives  at  the  end  of  its 
journey. 

Combining  equations  (45)  and  (46),  we  have 

s  =  1  (w  +  v)  t,  (47) 

showing  that  the  space  described  is  the  arithmetic  mean  of  ^U  and 
vt :  the  former  is  the  space  that  would  be  described  if  the  particle 
maintamed  its  origuial  velocity  u  through  the  whole  time  t ;  the 
latter  is  that  which  would  be  described  if  the  particle  had  its  final 
velocity  throughout  the  whole  time. 

Combining  equation  (47)   with  equation  (44),  which  can  be 

written  in  the  form 

ft  =  (r  -  u), 

we  obtain,  on  eliminating  t, 

2/s  =  v'  -  %l\  (48) 

an  equation  connecting  the  space  described  with  the  initial  and 
final  velocities. 

This  last  equation  may  also  be  deduced  from  the  equation  of 
energy.  Since  the  work  done  on  the  particle  is  equal  to  the  change 
in  its  kinetic  energy,  we  have 

Fs  =  ^  mv'^  —  \  inu^, 

and  since  P  =  mf,  equation  (48)  follows  at  once. 

Body  falling  under  Gravity 

155.  The  simplest  application  of  these  equations  is  to  the 
motion  of  a  body  wliicli  is  allowed  'to  fall  freely  under  the  influ- 
ence of  gravity,  so  that  the  acceleration  is  g. 

If  the  body  starts  from  rest,  we  put  u  =  0,  and  measure  s 
vertically  downwards.  We .  find  from  equation  (45)  that  after 
time  t  the  body  has   fallen  a  distance  Trfft',  while  its  velocity 


190 


MOTION  UNDEE  CONSTANT  FORCES 


time 


is  ft.    After  falling  a  distance  h  its  velocity  is,  by  equation  (48), 
equal  to  "v2^/?.    This  is  frequently  spoken  of  as  the  "velocity 

due  to  a  height  h." 

We  notice  that  the  distance 
fallen  varies  as  the  square  of  the 
time  during  which  the  body  has 
been  falling.  In  fig.  103  the 
time  is  measured  horizontally, 
while  the  distance  fallen  is 
measured  vertically.  The  thick 
curve  gives  a  graphical  repre- 
sentation of  the  distance  fallen. 
Denoting  the  horizontal  dis- 
tance by  X  and  the  vertical  by 
y,  we  have  x  =  t,y='^gf,&o  that 


distance  ^ 
fallen 


Fig.  10;5 


y^lgx". 


156.  This  is  the  equation  of  a  parabola,  so  that  the  curve  is  a 
parabola.  The  graph  can  be  obtained  experimentally  by  a  method 
known  as  Morin's  method.  A  weight 
P  is  free  to  fall  vertically  in  a  slot 
formed  in  a  rod  AB,  and  is  arranged 
so  that,  as  it  falls,  a  pencil  attached  to 
it  makes  a  mark  on  a  drum  CD  which 
is  covered  with  paper.  The  drum  is 
made  to  rotate  uniformly.  On  unroll- 
ing the  paper  from  the  drum  we  obtain 
the  graph  of  fig.  103,  —  for  the  hori- 
zontal distance  is  proportional  to  the 
time,  wliile  the  vertical  is  the  distance 
fallen  through.  The  fact  that  the  curve 
obtained  in  this  way  is  accurately  a 
parabola  gives  experimental  confirma- 
tion of  the  fact  that  motion  under  gravity  is  motion  with  uni- 
form acceleration. 


A  ^ 

p 

-^  ^ 



-^-== 

-^Z' 

2 

p    ^   - 



--_. 



i 

i    " 

Fig.  104 


BODY   FALLIInG  under  GRAVITY  191 

157.  If  the  body  is  projected  vertically  upwards  with  velocity 
u,  we  may  measure  the  distance  s  vertically  upwards,  and  the 
acceleration  m  this  direction  will  l)e  —  g.    Thus  we  have 

s  =  ut  —  I  gf, 

V  =  U  —  (jt, 

2  gs  =  tt^  -  1^, 


where  s  is  the  distance  upwards  described  after  time  t,  and  v  is 

the  upward  velocity.    From  the  first  equation  we  see  that  s  =  0 

2  1/ 
not  only  when  ^  =  0,  but  also  when  t  = Tlius  the  particle 

returns  to  its  original  position  after  time  2  u/g.  A\Qien  s  =  0  the 
tliird  equation  shows  that  tt^  =  v^.  Thus  wdien  the  particle  returns, 
its  velocity  is  the  same  as  when  it  started.  Clearly  this  must  be 
so,  for  the  potential  energy  is  the  same,  and  therefore  the  kinetic 
energy  also  is  the  same. 

EXAMPLES 

1.  If  an  express  train  is  doubled,  the  first  lialf  being  given  5  minutes'  start, 
and  attaining  its  maximum  booked  speed  of  48  miles  an  liour  after  moving  with 
constant  acceleration  for  a  mile,  prove  that  the  two  halves  will  lain  about  4 
miles  apart,  but  the  first  half  will  have  gone  3  miles  before  the  start  of  the 
second  half. 

2.  A  train  passes  another  on  a  parallel  track,  the  former  having  a  velocity 
of  45  miles  an  hour  and  an  acceleration  of  1  foot  per  second  per  second,  the 
latter  a  velocity  of  30  miles  an  hour  and  an  acceleration  of  2  feet  per  second 
per  second.  How  soon  will  the  second  be  abreast  of  the  first  again,  and  how 
far  will  the  trains  have  moved  in  the  meantime  ? 

3.  A  body  is  dropped  from  a  balloon  at  a  height  of  70  feet  from  the  ground. 
Find  its  velocity  on  reaching  the  ground,  if  the  balloon  is  (a)  rising,  (b)  falling, 
with  a  velocity  of  30  feet  a  second. 

4.  A  stone  is  dropped  into  a  "well,  and  the  sound  of  the  splash  reaches  the 
top  after  9  seconds.  Find  the  depth  of  the  well,  the  velocity  of  sound  being 
1100  feet  per  second. 

5.  An  elevator  descends  with  an  acceleration  of  5  feet  per  second  per  second 
until  its  velocity  is  20  feet  per  second,  after  which  its  velocity  remains  uniform. 
After  it  has  been  in  motion  for  6  seconds,  a  stone  is  dropped  on  to  it  from  the 
point  at  which  the  elevator  started.    How  soon  w'ill  it  strike  the  elevator  ? 

6.  A  juggler  keeps  three  balls  going  with  one  hand,  so  that  at  any  instant 
two  are  in  the  air  and  one  in  his  hand.  If  each  ball  rises  to  a  height  of  4  feet, 
show  that  the  time  during  which  a  ball  stays  in  his  hand  is  i  second. 


192 


MOTION  UNDER  CONSTANT  FORCES 


7.  A  body  was  observed  to  take  t  seconds  in  falling  past  a  hatchway  to  the 
bottom  of  a  hold  h  feet  deep.    Prove  that  it  fell 


and  struck  with  velocity 


-gi — \- -t\   feet, 

-  +  -gt  feet  per  second. 


8.  A  chain  12  feet  long  hangs  from  its  upper  end.  If  this  be  released,  find 
the  time  the  chain  will  take  in  passing  a  point  60  feet  below  the  initial  position 
of  the  highest  point. 

9.  A  body  whose  mass  is  5  pounds,  moving  with  a  speed  of  160  feet  per 
second,  suddenly  encounters  a  constant  resistance  equal  to  the  weight  of 
^  pound,  which  lasts  until  the  speed  is  reduced  to  96  feet  per  second.  For 
what  time  and  through  what  distance  has  the  resistance  acted  ? 

10.  Two  wagons,  coupled  together,  are  pulled  along  a  horizontal  track  by 
a  steady  force,  and  move  over  100  feet  in  the  first  ten  seconds  from  rest.  The 
rear  wagon  is  then  uncoupled,  and  it  is  found  that  at  the  end  of  the  next  ten 
seconds  the  interval  between  the  two  wagons  is  150  feet.  Compare  the  masses 
of  the  two  wagons,  all  resistance  being  neglected. 

11.  A  balloon  of  weight  W  is  rising  with  acceleration  /.  If  a  weight  w  of 
sand  be  emptied  out  of  the  car,  find  the  increase  in  the  acceleration  of  the 
balloon,  neglecting  the  resistance  of  the  air  and  the  buoyancy  of  the  sand. 


Motion  on  an  Inclined  Plane 

158.  Suppose  we    allow  a  particle  to  slide  do^\^l  an  inclined 
plane,    the  contact  between   the  two    being   supposed    perfectly 

smooth.  If  on  is  the  mass  of  the  par- 
ticle, the  forces  acting  on  it  are  its 
weight  7}ig,  and  the  reaction  E  normal 
to  the  plane.  The  component  down  the 
plane  is  mg  sin  a,  so  that  the  particle 
moves  with  uniform  acceleration  g  sin  a. 
We  can  obtain  the  distance  described 
in  the  time  t  from  the  usual  formulse. 

If  the  particle  starts  from  rest,  the  distance  described  in  time  t 

is  ^  g  sin  a  ■  f. 

159.  Suppose  that  through  a  point  0  we  have  a  great  number  of 
smooth  wires  on  which  smooth  beads  are  free  to  slide.     Let  tliese 


Fig.  105 


MOTIOX  ON  AN  INCLINED  PLANE 


193 


wires  make  all  possible  angles  with  the  vertical,  one  of  them,  00', 

being  vertical.    Let  us  imagine  that  the  beads  are  all  collected  at 

0  and   are   set  free   simultaneously. 

After  time  t,  let  the  bead  which  is 

falling  vertically  be  at  P,  and  let  the 

bead  which  is  falling  along   a  wire 

inclined  at  an  angle  /3  to  the  vertical 

be  at  Q.    This  latter  bead  moves  witli 

acceleration  g  cos  yS.    Thus  OP  =  1-  gt'-, 

while    OQ  —  \g  cos  yS  •  t"^.      Hence 

OQ==OP  cos  /3,  and  therefore  OQP 

is  a  right  angle.    It  follows  that  Q 

is  on  the  sphere  constructed  on  OP 

as  diameter,  and  obviously  the  same 

will    be   true    of   every   other   bead. 

Thus   at  any   instant  all  the  beads 

will  be  on  a  sphere  of  which  0  is  the  highest  point,  and  of  which 

the  lowest  point  is  at  a  distance  ^gf  below  0.    Hence  as  the 

motion  proceeds  the  beads  will  appear  to  form  a  sphere  which 

continually  swells  out  in  size,  the  highest  point  appearing  to 

remain  fixed  at  0,  while  the  lowest  point  appears  to  fall  freely 

under  gravity. 

160.  This  imaginary  experiment 
indicates  a  way  of  sohnng  a  prac- 
tical problem.  Suppose  we  wish 
to  place  a  smooth  plane  or  wire  in 
such  a  position  that  a  particle  will 
pass  do^vn  it  from  a  fixed  point  0 
to  a  given  fixed  surface  in  the  least 
time  possible.  Let  us  suppose  that 
we  fix  the  apparatus  of  wires  and 
beads  at  0,  that  we  set  the  beads 
free  simultaneously,  and  watch  the 
increase  in  size  of  the  sphere  which 
Fig.  107  they  form.    As  soon  as  the  sphere 


1U4 


MOTION  UNDEE  CONSTANT  FOECES 


reaches  such  a  size  that  it  touches  the  tixed  surface  at  some  point 
P,  one  of  the  beads  has  arrived  at  this  surface,  and,  moreover,  has 
arrived  in  shorter  time  than  any  of  the  others.  Thus  it  has  found 
the  quickest  path  from  0  to  the  surface.  This  path  is  OP,  and 
we  can  now  fix  the  path  without  performing  the  experiment,  from 
the  knowledge  that  a  sphere  drawn  so  as  to  have  0  as  its  highest 
point,  and  to  pass  through  P,  must  touch  the  surface  at  P. 

In  the  same  way,  if  we  wish  to 
find  the  quickest  time  from  a  surface 
to  a  fixed  point  0  below  it,  we  have 
to  find  a  sphere  winch  touches  the 
surface  at  some  point  P  and  has  0  for 
its  lowest  point.  Then  PO  will  be  the 
path  required.  For  it  is  seen  at  once 
that  the  time  down  all  the  chords  of 
this  sphere  which  pass  through  0  is 
the  same,  so  that  the  time  down  PO 
is  equal  to  the  time  down  any  other 
chord  QO,  and  therefore  less  than  the  time  dowTi  the  complete 
path  Q'O  from  the  surface  to  0,  of  which  the  chord  (>0  is  a  part. 


ILLUSTRATIVE   EXAMPLE 


A  ship  stands  some  distance  from  its  pier, 
and  it  is  required  to  place  a  chute  at  some  point 
of  the  ship''s  side,  so  that  the  time  of  sliding 
down  the  chute  on  to  the  pier  may  be  as  short 
as  possible. 

Clearly  the  lower  end  of  the  chute  must  just 
rest  on  the  nearest  point  0  of  the  pier,  and  the 
problem  reduces  to  that  of  drawing  a  sphere  to 
have  0  as  its  lowest  point  and  to  touch  the  ship's 
side.  Assuming  the  ship's  side  to  be  vertical, 
the  tangents  to  this  circle  at  the  ends  of  the 
chute  must  be  horizontal  and  vertical ;  whence  it 
is  easily  seen  that  the  chute  must  be  placed  so 
that  it  makes  an  angle  of  45°  with  the  vertical. 


Fig.  109 


AT  WOOD'S  MACHINE  195 

EXAMPLES 

1.  A  body  is  projected  with  a  velocity  of  20  feet  per  second  up  an  inclined 
plane  of  angle  45°.  Find  how  high  up  the  plane  it  will  go,  and  how  long  it  will 
take  in  going  up. 

2.  Two  particles  slide  down  the  two  faces  of  a  double  inclined  plane,  the 
angles  being  a  and  /3.  Compare  the  times  they  take  to  reach  the  bottom,  and 
the  velocities  they  acquire. 

3.  A  body  is  projected  down  an  inclined  plane  of  length  I  and  height  h,  from 
the  summit,  at  the  same  instant  as  another  is  let  fall  vertically  from  the  same 
point.  Prove  that  if  they  strike  the  base  at  the  same  time,  the  velocity  of 
projection  of  the  first  must  be 

1-'  -  fi^     IT 
I       \2h' 

4.  Give  a  construction  for  finding  the  line  of  quickest  descent  from  a  fixed 
point  to  a  circle  in  the  same  vertical  plane. 

5.  Particles  are  sliding  down  a  number  of  wires  which  meet  in  a  point,  all 
having  started  from  rest  simultaneously  at  this  point.  Prove  that  at  any  instant 
their  velocities  are  in  the  .same  ratio  as  the  distances  they  have  described. 

6.  A  railway  carriage  is  ob.served  to  run  with  a  uniform  velocity  of  10  miles 
an  hour  down  an  incline  of  1  in  250,  and  on  reaching  the  foot  of  the  incline 
runs  on  the  level.  Find  how  many  yards  it  will  run  before  coming  to  rest,  assum- 
ing the  resistance  to  be  constant  and  the  same  in  each  stage  of  the  motion. 

7.  Prove  that  if  a  motor  car  going  at  100  kilometers  an  hour  can  be  stopped 
in  200  meters,  the  brakes  can  hold  the  car  on  an  incline  of  about  1  in  5  ;  and 
determine  the  time  required  to  stop  the  car. 

8.  A  carriage  weighing  12  tons  becomes  uncoupled  from  a  train  which  is 
ranning  down  an  incline  of  1  in  250  at  a  rate  of  40  miles  per  hour.  The  fric- 
tional  resistance  is  14  pounds  weight  per  ton.  Find  how  far  the  carriage  will 
go  before  coming  to  rest. 

9.  The  pull  of  the  locomotive  exceeds  the  ordinary  resistances  to  the  motion 
of  a  train  by  -^^  of  its  whole  weight ;  and  when  the  brakes  are  full  on  there  Ls  a 
total  resistance  of  Jj  of  its  whole  weight.  Find  the  least  time  in  which  the 
train  could  travel  between  two  stopping  stations  on  the  level  3  miles  apart. 

10.  In  the  last  question,  find  the  time  if  the  track  is  down  a  gradient  of 
1  in  100. 

Atwood's  INIachine 

161.  It  is  difficult  to  measure  the  acceleration  produced  by 
gravity  from  direct  obser^^ations  on  a  body  falling  freely,  because 
either  the  distance  fallen  must  be  very  great  or  else  the  time 
of  faUing  very  small.  These  difficulties  are  to  some  extent  ob%dated 
in  a  machine  designed  by  Atwood. 


/ 


196  MOTION  UNDER  CONSTANT  FOECES 

If  a  string  having  two  equal  weights  attached  to  its  ends  is  placed 
over  a  smooth  vertical  pulley,  so  that  the  weights  hang  freely,  it  is 
obvious  that  there  will  be  equilibrium.  If  the  weights  are  unequal, 
equilibrium  cannot  exist.  In  Atwood's  machine  the  difference 
between  the  weights  is  made  small,  so  that  the 
motion  is  slow  and  is  therefore  easily  measured. 
Let  m^,  Wg  be  the  masses  of  the  weights, 
of  which  the  former  w^ill  be  supposed  to  be  the 
^'^  j,^,         greater.    When   set  free,  let  us  suppose  that 

j/    the  former  descends   with   an   acceleration  /. 
Regardmg  the  string  as  inextensible,  the  second 

mass  must  ascend  with  an  acceleration  /. 
lit* 

The  string  will  be  treated  as  weightless,  so 
Fig.  110  *=  . 

that  the  mass  of  any  element  of  it  may  be  dis- 
regarded. The  second  law  of  motion  accordingly  shows  that  the 
resultant  force  acting  on  any  element  must  vanish.  Thus  the  forces 
acting  on  the  string  must  be  in  equilibrium  (even  although  the 
string  is  not  at  rest),  and  it  follows,  as  in  §  54,  that  the  tension 
must  be  the  same  at  all  points,  say  T. 

The  forces  acting  on  either  mass  consist  of  its  weight  acting 
downwards  and  the  tension  of  the  string  acting  upwards.  Thus 
the  resultant  downward  forces  on  the  two  masses  are  respectively 
m^ff  —  T  and  m^j  —  T.  The  equations  of  motion  for  the  two  masses 
are  accordingly  ^^^^^  _  ^  =.  mj, 

m.^fj  —T  =  —  mj. 
If  we  eliminate  T,  we  oljtain 

giving  the  acceleration.    On  eliminating/,  we  obtain  as  the  value 
of  the  tension  ^ 

T= ^—^q.  (50) 

Clearly,  if  m^  is  nearly  equal  to  ih^,  the  acceleration  will  be  small.    For 
instance,  if  the  weights  are  100  and  101  grammes,  we  find  that 
f=  2^.9  =-016  feet  per  second  per  second. 


MOVING  FKA.ME  OF  REFERENCE  197 

An  acceleration  of  this  smallness  could  easily  be  measured ;  for  instance, 
the  heavier  mass  would  only  descend  eight  feet  in  ten  seconds.  In  prac- 
tice, the  difficulty  arises  that  if  the  difference  of  the  weights  is  made  t(jo 
small,  the  forces  acting  on  the  pulley  are  so  evenly  balanced  that  their 
difference  is  not  sufficient  to  overcome  the  friction  of  the  bearings,  etc. 

Motion  eeferred  to  a  Moving  Frame  of  Reference 

162.  It  has  already  been  seen  (§  25)  that  the  second  law  of 
motion  remains  true  when  the  motion  is  measured  relative  to  a 
frame  of  reference  which  is  not  at  rest  but  is  moving  with  a  uni- 
form velocity.  It  is  easy  to  find  how  the  statement  of  this  law 
must  be  modified  when  the  frame  of  reference  moves  with  a  known 
acceleration. 

Let  a  be  the  acceleration  of  the  frame  of  reference,  let  /  be  the 
component  of  the  acceleration  of  a  moving  particle  in  the  direction 
of  the  acceleration  a,  and  let  F  be  the  component  in  tliis  direction 
of  the  force  acting  on  the  particle.    By  the  second  law  of  motion 

P=m/',  (51) 

where  /'  is  the  component  acceleration  referred  to  a  frame  of  refer- 
ence at  rest.  The  acceleration/'  may,  however,  be  regarded  as  com- 
pounded of  the  acceleration  /  of  the  particle  relative  to  the  moving 
frame  of  reference,  together  with  the  acceleration  a  of  this  frame 
relative  to  one  at  rest.  Since  these  accelerations  are  all  in  the 
same  direction,  we  have  f  =f  +  a,  so  that  equation  (51)  becomes 

p  =  m  (/  +  a). 

We  can  also  write  this  in  the  form 

P  —  ma  =  mf,  (52) 

showing  that  tJie  motion  is  the  same  as  if  the  frame  were  at  rest, 
provided  we  imagine  the  force  P  diminished  hy  an  amount  ma. 

This  result  can  easily  be  interpreted  pliysically.  Of  the  force  P, 
a  part  equal  to  ma  is  used  up  in  causmg  the  particle  to  keep  pace 
with  the  moving  frame  of  reference.  It  is  only  the  remaining  part, 
P  —  ma,  which  is  available  for  producing  accelerations  relative  to 
the  moving  frame. 


198 


MOTION  UNDER  CONSTANT  FORCES 


163.  Frame  moving  with  vertical  acceleration.  If  the  frame 
of  reference  moves  with  an  acceleration  a  vertically  downwards, 
we  see  that  before  measuring  accelerations  relative  to  this  frame, 
we  must  suppose  the  vertical  component  of  force  on  each  particle 
of  mass  m  to  be  dimhiished  by  ma.  Whatever  forces  are  acting, 
there  will  be  amongst  them  the  weights  of  the  particles  mg,  etc. 
We  can  conveniently  suppose  the  diminution  7na  to  be  taken 
from  these,  so  that  the  weight  of  a  particle,  instead  of  being 
taken  to  be  mg,  will  be  taken  to  be  7n  (g  —  a). 

Thus  the  acceleration  of  the  frame  of  reference  may  be  allowed 
for  by  supposing  the  acceleration  due  to  gravity  to  be  diminished 
from  g  to  g  —  a. 

For  example,  if  an  Atwood's  machine  is  jilaced  in  an  elevator,  then  at 
the  instant  at  which  the  elevator  has  an  upward  acceleration  a,  the  accel- 
eration of  the  masses  relative  to  the  machine  will  be  (cf.  equation  (49)) 


/  = 


(,'/+«)» 


nil  +  /«., 
while  the  tension  of  the  string  will  be  (cf.  equation  (50)) 


/III  +  III. 


((/  +  «)• 


164. 

seen  ( 


V' 


a  cos\ 


Effect  of  earth's  rotation  on  the  value  of  g.    As  we  have 

25),  a  frame  of  reference  which  is  fixed  relatively  to  the 
earth's  surface  possesses  an  acceleration 
in  consequence  of  the  rotation  of  the 
earth  about  its  axis. 

Let  00'  be  the  earth's  axis,  and  let 
P  1)6  any  point  on  the  earth's  surface, 
in  latitude  X.  Regarding  the  earth  as  a 
sphere  of  radius  a,  the  point  F  will  de- 
scrilae  a  circle  of  radius  a  cos  \  having 
its  center  N  on  the  earth's  axis.  If  v  is 
the  velocity  with  which  P  describes  the 
circle,  the  acceleration  of  P  is,  by  §  12, 

towards  the  center  of  the  circle,  —  i.e.  along  PA^. 


MOYIXG   FEAME  OF  REFEEEXCE  199 

Let  a>  be  the  angular  velocity  of  the  earth,  —  i.e.  let  it  turn  through 
<sy  radians  per  unit  time.  Then  the  time  of  P  describing  a  com- 
plete circle  is  the  same  as  the  time  required  for  the  earth  to  perform 

,     .                1     -i  TT      „,  .      .        .      1      2  TT  a  cos  \ 
a  complete  revolution,  namely ihis  time  is  also 

Hence  we  have  ^ 

-y  =  ao)  cos  A. 

The  acceleration  of  the  frame  of  reference  is  now  seen  to  be 

=  (ip-a  cos  \ 

a  cos  X 

along  FN.  The  motion  of  any  particle  referred  to  a  frame  moving 
with  F  may  accordingly  be  calculated  as  though  with  reference  to 
a  fixed  frame,  provided  the  component  of  force  in  the  direction 
FN  is  diminished  by  iiKtP'a  cos  X. 

Thus  the  total  force  acting  may  be  supposed  to  consist  of  the 
forces  which  actually  do  act,  combined  with  a  force  maP'a  cos  \ 
along  NF.  Compounding  this  last  force  with  the  earth's  attrac- 
tion, we  obtain  a  force  which  may  be  called  the  apparent  force  of 
gravity  at  F.  Thus  the  motion  of  the  frame  of  reference  may  be 
allowed  for  by  using  the  apparent  force  of  gravity  in  place  of  the 
true  attraction  of  the  earth.  It  is  this  apparent  gravity  which  is 
determined  experimentally,  and  which  is  always  meant  in  speaking 
of  the  iveight  of  a  particle  at  any  point. 

To  find  the  apparent  weight  of  a 
l)ody  at  the  point  F,  we  have  to  com- 
pound its  true  weight,  say  inG  acting 
along  FC,  with  a  force  moi'a  cos X,  along 
NF.  Let  the  latter  force  be  resolved 
into  its  components  Fig.  112 

—  mora  cos^  X,     mat^a  cos  X  sui  X 
along  FC,  FT  respectively,  FT  being  the  tangent  at  F. 

Compounding  with  the  force  mG  along  FC,  we  find  for  the  com- 
ponents X,  Y  of  the  apparent  weight  along  FC,  FT  respectively, 
X  =  111  (G  —  fo^a  COS"  X),  (53) 

Y  =  mco^  a  cos  X  sin  X.  (54) 


DR.  GEORGE  F.  McEWEH 

200  MOTION   U^^DER  CONSTANT  FORCES 

Squaring  and  adding,  anil  denoting  the  apparent  weight,  as  usual, 
by  m(j,  we  obtain 

wy  =  X-  +  Y^  =  nv" (G-  -  2  (o\i  G  cos'  X  +  co'a^  cos^  X).      (55) 

Taking  the  diameter  of  the  earth  to  be  7927  miles,  and  the 

value  of  G  (the  acceleration  due  to  gravity  at  the  North  Pole) 

to  be  32.25,  we  easily  find  that 

ft)^a         1  ■      ,  1 
=  ^'  approximately. 

The  square  of  this  is  so  small  that  to  a  first  approximation  it 
may  be  neglected,  and  equation  (55)  may  be  written  in  the  form 
(J  =  G  —  ay^a  cos'  X. 

Thus  the  apparent  weight  in  latitude  X  is  less  than  the  true 
weight  by  may^a  cos'  X,  or  about  gg-Q  cos'  X  of  the  whole  weight. 

The  apparent  weight  does  not  act  along  the  radius  CP.  If  we 
suppose  it  to  act  at  an  angle  6  with  this  radiuS,  we  obtain,  from 
equations  (53)  and  (54), 

^       Y      co'a  cos  X  sin  X 

tan  U  =  —  = -— — -— 

X      G  —  (o  a  cos  X 

=  o-^-Q  ^°^  ^  ^^^^  ^'  approximately, 

giving  the  deviation  of  the  plumb  line  from  the  earth's  radius  at 

any  point. 

Frictional  Eeactions  between  Moving  Bodies 
165.  It  is  found  experimentally  that  the  relation 

(in  which  F,  R  are  the  tangential  and  normal  components  of  the 
reaction  between  two  bodies)  remains  very  approximately  true 
when  the  bodies  are  sliding  past  one  another.  The  value  of  /^t, 
the  coefficient  of  friction,  is  not  quite  the  same  as  wlien  the 
bodies  are  at  rest,  the  latter  being  always  somewhat  larger. 

Friction  between  two  bodies  which  are  sliding  past  one  another 
is  called  dynamical  friction,  that  between  two  bodies  at  rest  being 
called  statical  friction. 


ILLUSTKATIVE   EXAMPLES 


201 


ILLUSTRATIVE  EXAMPLES 

1.  Tioo  particles  of  masses  nii,  mo  are  placed  on  two  inclined  planes  of  angles 
a,  ^,  placed  back  to  back,  and  are  connected  by  a  string  which  passes  over  a  smooth 
pulley  at  the  top  of  the  planes.  If  the  coefficients  of  friction  between  the  particles 
and  the  planes  are  /mi,  ij.2,  find  the  resulting  motion. 

If  motion  occurs  at  all,  one  particle,  say  mi,  must  move  down  its  plane,  while 
the  other,  m-z,  will  move  up.  Since  the  string  is  inextensible,  the  acceleration 
of  each  will  be  the  same,  say /in  the  direction  in  which  motion  is  taking  place. 

The  forces  acting  on  the 
first  particle  are 

(a)  its  weight  m\g  ver- 
tically down  ; 

(6)  the  tension  of  the 
string,  say  T,  up  the  plane  ; 

(c)  the  reaction  with 
the  plane.  Let  this  be  re- 
solved into  components  i?, 
yiR  normal  to  and  up  the 
plane. 

Since  the  particle  mi  has  no  acceleration  normal  to  the  plane,  the  component 

of  the  resultant  force  in  this  direction  must  be  zero.    Resolving  in  this  direction 

we  obtain 

R  —  mig  coscr  =  0. 

Resolving  down  the  plane, 

mig  sin  a  —  /xR  —  T  =  mif, 

and  if  we  eliminate  the  unknown  reaction  R,  we  obtain 


Fig.  113 


mig  (sin  a  —  fi  cos  a)  —  T  =  mif. 


(a) 


A  similar  equation  can  be  obtained  for  the  motion  of  the  second  particle, 

namely 

niog  (sin (3  -|-  fx  cos/3)  —  T  =  —  niof  {b) 


Solving  equations  (a)  and  (5)  for/,  we  obtain 

_  mi  (sin  a  —  fjL  cos  a)  —  mo  (sin/3  -f  fi  cos/3) 


('') 


mi  +  m^ 
giving  the  acceleration. 

If  this  value  of /comes  out  negative,  we  see  that  the  acceleration  cannot  be 
in  the  direction  in  which  motion  has  been  assumed  to  take  place. 

If  the  system  starts  from  rest,  motion  in  the  direction  assumed  is  found  to 
be  impossible,  and  we  must  proceed  to  examine  wdiether  motion  in  the  opposite 
direction  is  possible.  If  this  also  is  found  to  be  impossible,  the  system  will 
remain  at  rest. 


202 


MOTION  UNDER  CONSTANT  FORCES 


If,  however,  the  system  is  known  to  have  been  started  in  motion  in  the  direc- 
tion assumed,  then  the  acceleration  given  by  equation  (c)  will  be  in  operation, 
increasing  the  velocity  if  positive,  and  decreasing  it  if  negative.  In  the  latter 
case  the  system  will  in  time  be  reduced  to  rest,  and  we  must  then  examine 
whether  or  iiot  it  will  start  into  motion  in  the  reverse  direction. 


2.  To  one  end  of  the  string  of  an  Atwood's  machine  a  weight  of  mass  nii  is 
attached.  To  the  other  end  a  smooth  pulley  of  mass  m^  is  attached,  over  which 
passes  a  string  with  masses  wis,  mi  hanging  at  its  ends.    Find  the  motion. 

Let  the  mass  mi  be  supposed  to  have  an  acceleration  /,  measured  downwards. 
Then  m^  must  have  an  acceleration  /  upwards.    The  masses  ms,  mi  will  them- 
selves form  an  Atwood's  machine,  the  whole  of  which 
moves  upwards  with  an  acceleration/.    Thus  the  ten- 
sion in  the  string  of  this  machine,  say  Ti,  is  (cf.  §  163) 


J 


Ti 


2  m^nii 
ma  +  Mi 


(9+f)- 


(«) 


r^- 


If  we  denote  the  tension  in  the  string  connecting  mi 
and  7?i2  by  T^,  we  have  as  the  equation  of  motion  of  m^, 


Fig.  114 


To  _  m.sr  -  2  Ti  =  m^f, 
while  the  equation  of  motion  of  lui  is 
■mig  -  T.2  =  mi/. 


(&) 


(c) 


Eliminating  Ti  and   T2  from  equations  (a),  (b),  and  (c),  we  obtain  as  the 
value  of  the  acceleration/, 

97?  1  —  7/(2 


4  77l3?7l4 


/  = 


77^:5  +  m4 


77*1  +  7/(2 


47/(37774 
7ft3  +  IHi 


The  accelerations  of  the  masses  7n3,  7774  relative  to  7772  are  known,  by  §  163, 
to  be 


ms  —  m4 
ms  +  ?/i4 


(g+f)- 


3.  At  equal  interval^  on  a  horizontal  circle  n  small  smooth  rings  are  fixed, 
and  an  endless  string  passes  through  them  in  order.  If  the  loops  of  the  string 
between  each  consecutive  pair  of  rings  support  n  pulleys  of  masses  P,  Q,  R,  •  •  • 
respectively,  the  portions  of  string  not  in  contact  with  the  pulleys  being  vertical, 
show  that  the  pulley  P  will  descend  with  acceleration 

1-n       1        1 


1  1  1 


ILLU8TEAT1VE  EXA:MPLES 


203 


The  tension  of  the  string  must  be  the  same  tliroughout,  say  T.  If  the  accel- 
erations of  the  pulleys  are  //^,  fq,  ■  ■  ■ ,  all  measured  down,  we  have  ecjuations 
of  motion  of  the  type 

Pg  -2T=Pfp, 

one  equation  for  each  pulley.  The  unknown 
quantity  T  enters  these  equations,  as  well  as 
the  n  unknown  quantities  //>,  /q,  •  ■  • .  Thus 
there  are  n  +  1  unknown  quantities,  and  so 
far  only  n  equations  connecting  them.  An- 
other equation  is  therefore  required,  and  this 
is  obtained  by  noticing  that  the  accelerations 
/p,  /q,  •  •  •  cannot  be  independent,  for  the 
length  of  the  string  must  remain  unaltered. 

Let  us  denote  the  depths  of  P,  Q,  •  •  •  below  the  horizontal  ring  by  Sp,  Sq, 

Then  sp  -\-  sq  +  ■  ■■ 

must  be  constant  throughout  the  motion.    It  follows  that 


fp+fQ  + 


0. 


Substituting  the  values  of /p, /q,  •  •  •  from  equation  (a),  we  obtain 
so  that  2  T  : 


Q 
ng 

1       1 

P+Q  + 


and  on  substituting  this  value  for  2  T  in  equation  (a),  we  obtain  the  required 
value  of  /p. 

EXAMPLES 

1.  Show  that  the  tension  of  the  string  in  an  Atwood's  machine  is  Intermediate 
between  the  weights  of  the  two  masses.  Show  also  that  it  is  nearer  to  the 
smaller  than  to  the  larger  of  these  weights. 

2.  Two  weights  16  and  14  ounces  respectively  are  connected  by  a  light  inex- 
tensible  string  which  passes  over  a  smooth  pulley.  The  weights  hang  with  the 
strings  vertical  and  the  string  is  clamped  so  that  no  motion  can  take  place.  If 
the  string  is  suddenly  undamped  find  the  change  in  the  pressure  exerted  on  the 
pulley. 

3.  A  string  passing  across  a  smooth  table  at  right  angles  to  two  opposite 
edges  has  attached  to  it  at  the  ends  two  masses  P,  Q  which  hang  vertically, 
f  rove  that,  if  a  mass  M  be  attached  to  the  portion  of  the  string  which  is  on  the 
table,  the  acceleration  of  the  system  when  left  to  itself  will  be 

P-Q 

P+Q  +  M^' 


204  MUTiOX   UNDER  CONST.iNT  FORCES 

4.  Two  masses  ?«,  in'  are  tied  to  the  two  ends  of  a  string  which  is  shing  over 
a  peg,  as  in  an  Atwood's  machine.  The  peg  is  not  smooth,  the  angle  of  friction 
between  it  and  tlie  string  being  e.    Find  tlie  motion. 

5.  In  question  3,  let  the  coefficient  of  friction  between  the  table  and  tlie 
weight  M  be  fi,  that  between  the  table  and  string  being  /x'.    Find  the  motion. 

6.  A  rope  hangs  over  a  smooth  pulley.  Find  the  uniform  acceleration 
with  which  a  man  weighing  10  stone  must  pull  himself  up  one  end  of  the 
rope,  for  the  rope  to  be  kept  at  rest  by  a  weight  of  12  stone  hanging  from 
the  other  end. 

7.  A  monkey  is  tied  to  one  end  of  the  string  of  an  Atwood's,  machine,  the 
other  end  having  attached  to  it  a  weight  exactly  equal  to  that  of  the  monkey, 
and  at  just  the  same  depth  below  the  pulley.  The  monkey  suddenly  starts  to 
climb  up  his  string.  Which  will  rise  the  faster,  the  monkey  or  the  weight  on 
the  other  string  ? 

8.  Weights  of  10  pounds  and  2  pounds,  hanging  by  vertical  strings,  balance 
on  a  wheel  and  axle.  If  a  mass  of  1  pound  be  added  to  the  smaller  weight,  find 
the  acceleration  with  which  it  will  begin  to  descend,  and  the  tension  of  each 
rope.    (The  inertia  of  the  wheel  and  axle  is  to  be  neglected.) 

9.  A  mass  of  5  pounds  resting  on  a  smooth  plane  inclined  at  30  degrees  to 
the  horizon  is  connected  by  a  fine  thread,  which  passes  over  a  pulley  at  the 
summit  of  the  plane,  with  a  mass  of  3  pounds  hanging  vertically.  Compare  the 
pull  in  the  thread  when  the  mass  on  the  plane  is  held  fixed  and  when  it  is  let 
go.  If  the  thread  is  severed  or  burnt  8  seconds  after  this  mass  has  been  let  go, 
find  how  far  it  will  rise  on  the  plane  before  falling  back. 

10.  A  light  thread  passes  over  two  fixed  pulleys  A  and  B,  and  carries 
between  them  a  movable  pulley  block  C,  under  which  it  passes.  A  mass  M  is 
attached  to  each  end  of  the  thread,  and  a  mass  m  to  the  movable  block.  The 
masses  of  the  pulleys  are  negligible,  and  the  pulleys  are  so  arranged  that  all 
the  segments  of  the  thread  are  vertical.  Show  that,  when  the  system  is  let  go, 
the  tension  in  the  thread  is  mM/{M  +  h  m)  pounds,  and  find  the  acceleration 
with  which  the  mass  m  falls. 

11.  An  elastic  band  of  mass  m,  natural  length  a,  and  modulus  X  is  placed 
round  a  rough  horizontal  wheel  of  circumference  b  (>  a).  How  fast  must  the 
wheel  be  made  to  rotate  for  the  band  to  leave  the  wheel  ? 

12.  The  elastic  band  of  question  11  is  placed  on  a  smooth  sphere  of  circum- 
ference b  rotating  with  angular  velocity  w.    Find  the  position  of  rest. 

13.  If  the  earth  rotates  faster  and  faster  until  ultimately  bodies  fly  off  from 
its  equator,  show  that  by  the  time  this  stage  is  reached  the  plumb  line  at  any 
point  will  be  parallel  to  the  earth's  axis. 

14.  A  body  is  placed  on  a  spring  balance  when  in  a  ship  which  is  sailing 

along  the  equator  with  velocity  v.    Show  that  if  the  balance  weighs  accurately 

when  the  ship  is  at  rest,  its  reading  when  the  ship  is  in  motion  will  show  an 

.  2vo) 
error  of  times  the  weight  of  the  body  (approximately),  where  w  is  the 

angular  velocity  of  the  earth. 


PEOJECTILES  205 

Flight  of  Peojectiles 

166.  By  a  projectile  here  is  meant  any  body  which  is  Small 
enough  to  be  regarded  as  a  particle,  and  which  is  projected  in  such 
a  way  that  it  describes  a  path  under  the  influence  of  gravity. 

A  projectile  will,  in  general,  be  influenced  by  the  resistance  of 
the  air  as  well  as  by  gravity,  but  we  shall  suppose  the  resistance 
of  the  air  to  be  negligible,  so  that  gravity  will  be  the  only  force 
wliich  need  be  taken  into  account. 

To  take  the  simplest  case  first,  let  us  imagine  that  the  projectile 
is  projected  horizontally  from  the  point  0  (fig.  116),  with  velocity  w. 
The  only  force  acting  is  gravity,  which  q 
has  no  horizontal  component,  so  that 
the  horizontal  velocity  remains  equal  to 
%L  throughout  the  motion.  The  initial 
vertical  component  of  velocity  is  nil, 
but  there  is  a  dowTiward  acceleration  g. 
Thus  after  time  t,  the  horizontal  dis- 
tance described  is  ut,  while  the  vertical  y^ 
distance  fallen  is  \gf.  Denoting  the 
horizontal  distance  described  by  x,  and  the  vertical  distance  fallen 


Fig.  IKi 


by  y,  we  have 


X  =  ut,     y  =\gt 


The  equation  of  the  path  described  is  obtained  by  eliminating  t 
from  these  equations,  and  it  is  found  to  be 

Tins  is  a  parabola,  of  which  the  latus  rectum  is • 

9 

Clearly  the  problem  of  determining  the  curve  is  essentially  the  same  as 
in  §  156.  There  we  have  a  body  falling  freely,  and  tracing  its  path  on  a 
paper  vMcli  moves  past  it  with  a  uniform  horizontal  velocity.  Here  we  have 
a  body  falling  freely,  and  can  imagine  it  to  trace  its  path  on  a  paper /josf 
ichich  it  moves  with  a  uniform  horizontal  velocity.  The  relative  motion  is 
the  same  in  the  two  cases,  so  that  the  curves  are  necessarily  the  same. 


206 


MOTION  UNDER  CONSTANT  FORCES 


167.  At  0  the  velocity  of  the  particle  is  u,  wliich  is  the  velocity 

due  to  a  height  - —    Tliis  is  equal  to  a  quarter  of  the  latus  rectum, 

and  is  therefore  equal  to  the  depth  of  0,  the  vertex  of  the  parabola, 
below  the  directrix  XM.     Thus  the  total  energy  of  the  projectile 

when  at  0  is  equal  to  that  of  the  same 
projectile  at  rest  at  X,  or,  of  course,  at 
any  other  point  of  the  directrix,  since 
this  is  horizontal. 

Since  tlie  total  energy  remains  con- 
stant, we  see  that  when  the  particle  is 
at  any  point  P  of  its  path,  its  kinetic 
energy  is  that  due  to  a  fall  through  PJf, 
^^°"  -^^^  the  distance  from  P  to  the  directrix. 

This  is  expressed  by  saymg  that 

The  velocity  of  a  projectile  at  any  point  is  that  due  to  a  fall 
from  the  directrix. 

168.  Instead  of  supposing  that  the  particle  is  projected  horizon- 
tally at  0,  the  vertex  of  the  parabola,  we  can  suppose  that  it  has 
arrived  at  0  in  its  flight  through  the  air,  having  been  pre\'iously 
projected  from  some  point  A.  The  same  reasoning  which  shows 
that  the  part  of  the  path  described  ^ 
after  passing  0  is  parabolic,  will 
show  that  the  path  described  be- 
fore reacliing  0  is  paraboHc  also. 
Thus  the  path  of  a  particle  pro- 
jected from  any  point  in  any 
manner  is  a,  parabola. 

Suppose  that  a  particle  is  pro- 
jected from  A  with  velocity  v,  in 
a  direction  wliich  makes  an  angle 
a  with  the  horizontal.  Let  0  be  the  vertex  of  its  path,  and  let  us 
suppose  that  when  the  projectile  passes  through  0,  its  velocity  is 
u,  this  velocitv  being  of  course  horizontal. 


PROJECTILES  207 

There  is  no  horizontal  force  acting  on  the  particle,  so  that  its 
horizontal  velocity  remains  unaltered  throughout  its  flight.    Thus 

u  =  V  cos  a. 
The  latus  rectum  of  the  parabola  is  accordingly 

2  u^      2  v"^  cos*^  a 
9    ~         9 

The  velocity  v  is  that  due  to  a  fall  from  the  directrix  to  A,  so 
that  if  NX  is  the  directrix  in  fig.  118, 

^  AN=^. 

The  time  of  fliglit  from  ^  to  0  is  the  time  required  for  gravity 

V  sin  (t 
to  destroy  a  vertical  velocity  v  sin  a ;  it  is  therefore In 

9 
this  time  the  horizontal  distance  A3I  is  described  with  a  uniform 

horizontal  velocity  ti,  so  that 

V  sin  a         v^  sin  a  cos  a 

AM  = to  = 

9  9 

The  vertical  distance  described,  031,  is  by  equation  (47)  equal 
to  half  of  the  time  multiplied  by  the  initial  vertical  velocity. 
Thus  -,    o    .  o 

2        9 
The  total  range  on  a  horizontal  plane,  AA',  is  twice  AM,  so  that 
2  v^  sin  a  cos  a      v^  sin  2  a 


AA'  = 


9  9 


169.  If  the  value  of  v  is  fixed  (as,  for  instance,  it  would  be  if 
we  were  firing  a  shot  with  a  given  charge  of  powder),  while  the 

angle  a  can  be  varied,  then  the  range  A  A'  can  never  exceed  —>  for 

9 
the  factor  sin  2  a  can  never  exceed  unity.    Thus  the  greatest  range 

attainable  on  a  horizontal  plane  with  a  given  velocity  of  projection 


208  MOTION  UNDER  CONSTANT  FORCES 

o 

V 

V  will  be  —J  and  to  obtain  this  range  we  make  sin  2  a  =  1,  or 
9 

a  =  45  degrees.    Thus,  to  send  a  projectile  as  far  as  possible  on 

a  horizontal  plane  we  project  it  at  an  angle  of  45  degrees. 

170.  These  results  can  also  be  obtained  analytically.    Let  us 

take  the  point  of  projection  for  origin,  and  the  plane  in  which 
the  flight  takes  place  as  plane  of  xj/,  the  axes  of 
X  and  y  being  respectively  horizontal  and  vertical. 
The  «-co6rdmate  of  the  point  reached  by  the 
particle  after  time  t  is  e({ual  to  the  horizontal 
distance  described  in  time  t  with  uniform  hori- 
zontal velocity  v  cos  a.    Thus 

Fig.  119 

X  =  V  COS  a  •  t.  (56) 

Similarly  the  y-coordinate  of  this  point  is  the  distance  described 
in  time  t,  starting  with  initial  velocity  v  sin  a,  and  with  retarda- 
tion q.    Thus  .  ,      „  ,__, 
^                               y  =  v  sin  a  •  t  —  \gf.                                   (57) 

If  we  eliminate  t  between  equations  (56)  and  (57),  we  obtain 
the  equation  of  the  path.    It  is  found  to  be 

1/  =  X  tana ^ -— .  (58) 

This  can  be  expressed  m  the  form 

1  r  sm"  a  <l         I        v^  sin  a  cos  a 


g  2v^  cos^  a\  g 

which  is  clearly  the  equation  of  a  parabola,  of  which  the  vertex  is 

at  the  point 

■?'^  sin  a  cos  a               Ir'^sm^a  .__, 

x  = ,     y  =  - ,  (59) 

//  ^        9 

and  of  wliich  the  latus  rectum  is  of  length 

2  v^  cos^  a 
9 


PPvOJECTILES 


209 


To  obtain  the  range  on  a  horizontal  plane,  we  have  to  find  the 
point  in  which  the  parabola  intersects  the  Ihie  y  =  0.  Putting 
?/  =  0  in  equation  (58),  we  obtain  at  once 

2  v"^  cos-  a  v^  sin  2  a 

X  = tan  a  = > 

9  9 

agreeing  with  the  value  obtained  in  §  168. 


Range  on  an  Inclined  Plane 

171.  Suppose,  next,  that  the  projectile  is  fired  so  as  to  strike 
an  inclmed  plane  through  0,  the  point  of  projection.  Let  ^  be 
the  inclination  of  this  plane  to  the 
horizon,  and  let  r  be  the  range  of  the  y 
projectile  on  this  plane.  Then  the  co- 
ordiuate  of  the  point  at  which  the 
jDrojectile  meets  the  plane  must  be 

X  =  r  cos  /3,     y  —  T  sin  yS. 

Tins  point  is  a  point  on  the  parab- 
ola, so  that  its  coordinates  must  sat-    ^ 
isfy  equation  (58).    Substitutmg  these 
coordinates,  we  obtain 


Fig.  120 


r  sin  ^  =  r  tan  a  cos  /3 
giving  as  the  value  of  the  range  r, 


g  r^  cos^  /3 
2  v^  cos''^  a 


r  = 


2  v^  cos  a  sin  (a  —  /3) 


Since 


g  cos''^  ^ 

%  cos  a  sm  (a  —  /3)  =  sin  (2  a  —  /3)—  sin  /3, 


(60) 
(61) 


it  is  clear  that  if  a  alone  is  allowed  to  vary,  the  range  r  will  be  a 
maximum  when  sin  (2  a;  —  /S)  is  a  maximum,  i.e.  when  it  is  equal 
to  unity.    To  obtain  tliis  value,  we  make 


a  = 


•i 


210 


MOTION  UNDER  CONSTANT  FORCES 


Thus,  to  get  the  maximum  range,  we  project  in  the  direction  which 
bisects  the  angle  between  the  inclined  plane  and  the  vertical. 

When  projection  takes  place  in  this  direction,  the  maximum 
range  R  is  given  by  putting  sin  (2  «  —  ^S)  =  1  in  the  value  for  r 
given  by  equation  (60).    Thus  we  have 

_  t-^  2  cos  a  sin  {a  —  /3) 
g  cos^  /3 

v^  sin  (2  a:  —  /3)  —  sin  yS 
g  COS^yS 

v^  1  —  sin  /3 


y 


cos'^  /3 


(62) 
^(l  +  sin/8)  ^     ^ 

172.  This  equation  enables   us  to   find  the  greatest  distance 
which  can  be  reached  in  any  direction  by  a  projectile  fired  with 


Let  us  replace  /3  by  —  —  ^,  so  that  6  is  the  angle 


velocity  v. 

which  the  direction  makes  wdth  the  vertical.    Then  the  relation 
between  B  and  Q  is 


B  =  -—-^ ^'  (63) 

^(1  +  cos^)  ^     ' 

Regarded  as  an  equation  in  polar  coordinates  E,  6,  this  is  clearly 
the  equation  of  a  curve  such  that  we  can  Mt  any  point  inside  it 

with  a  projectile  fired  with 
velocity  v,  but  cannot  reach 
any  point  outside  it.  The 
polar  equation  of  a  parabola 
of  latus  rectum  /,  referred 
to  its  focus  and  axis,  is 
known  to  be* 


E  = 


1  +  cos  0 


Fig.  121 

Comparing  this  with  equation  (63),  we  see  that  this  equation 
represents  a  parabola,  of  which  the  point  of  projection  is  the 
focus,  the  axis  is  vertical,  and  the  semi-latus  rectum  is  v^/g. 


PKOJECTILES 


211 


Envelope  of  Paths 

173.  If  we  imagine  all  the  parabolas  drawn,  which  can  be 
described  by  projectiles  fired  from  the  point  0  with  a  given 
velocity  v,  we  shall  obtain  a  figure  similar  to  fig.  122.  The  out- 
side curve  obviously  separates  the  points  which  can  be  reached 


Fig.  122 


from  those  wliich  cannot  be  reached.  Thus  this  is  the  parabola  of 
which  the  equation  is  given  in  equation  (63).  A  study  of  fig.  122 
will  now  show  that  this  curve  is  the  envelope  of  the  system  of 
parabolas  which  correspond  to  the  different  directions  of  firing. 

174.  The  envelope  of  the  system  of  parabolas  can  be  found  more 
directly  by  analytical  methods.  If  we  write  m  for  tan  a  in  equa- 
tion (58),  we  obtain  the  equation  of  a  parabola  of  the  system  in 
the  form 

2  v' 


y  =  mx 


.(1  +  m^), 


and  the  whole  system  is  obtained  by  giving  different  values  to  m. 
The  condition  for  this  equation  to  have  equal  roots  in  m  is  that 


or,  in  reduced  form, 


^2^4^ 


2v^ 


U 


f^-^ 


■"-2^ 


(64) 


If  X,  y  satisfy  this  relation,  two  parabolas   which  only  differ 
infinitesimally  pass  through  x,  y,  and  therefore  x,  y  is  a  point  on 


212 


MOTION  UNDER  CONSTANT  FORCES 


the  envelope.  Thus  equation  (64)  is  the  equation  of  the  envelope, 
and  is  easily  seen  to  give  the  same  parabolic  envelope  as  has  already 
been  obtained. 

175.  There  is  also  a  very  simple  geometrical  way  of  deter- 
mining the  envelope  of  the  system  of  parabolas.  We  notice  first 
that  as  the  projectiles  are  all  tired  from  the  same  point  A  with 
the  same  velocity  v,  their  paths  must  all  have  the  same  directrix 
^'M  (tig.  123). 

Let  any  two  parabolas  of  the  system  hitersect  in  F,  and  let 
*S',  *S"  be  the  foci  of  these  parabolas.    Let  AX,  PM  be  the  perpen- 
diculars from  A  and  F  to 
the  directrix. 

Then  AS  =  AS',  since 
each  is  equal  to  AN,  and 
FS  =  FS',  for  each  is 
equal  to  F3f.  Thus  S,  S' 
are  the  two  points  of  in- 
tersection of  two  circles  of 
which  the  centers  are  A,  F. 
If  the  two  parabolas  are 
supposed  to  be  adjacent, 
their  foci  S,  S'  are  adjacent  points,  and  therefore  the  two  circles 
touch,  and  ASF  is,  in  the  limit,  a  straight  line.    We  now  have 

AF  =  AS  +  SF 
=AN+FM 

=  the  perpendicular  from  P  on  to  a  fixed  horizontal 
line  at  a  distance  AX  above  MX. 

Tlie  point  F,  then,  satisfies  the  condition  that  its  distance  from 
this  fixed  line  is  equal  to  its  distance  from  the  fixed  point  A.  It 
therefore  is  always  on  a  certain  parabola  of  focus  A.  But  also  it  is 
always  a  point  on  the  envelope,  this  being  the  locus  of  the  points 
of  intersection  of  adjacent  pairs  of  the  parabolas  of  the  system. 
Thus  the  envelope  is  the  parabola  just  obtained,  of  which  the  focus 
is  A,  and  this  is  tlie  same  parabola  as  was  obtained  before. 


Fig.  123 


ILLUSTKATIVE   EXAMPLES 


213 


ILLUSTRATIVE   EXAMPLES 


iiV  =  iP  +  PiV  =  a  (1  +  cos  2  0), 

while  the  vertical  component  of  its  velocity  is 

V  •  iQL/a)  sin  ^  =  2  F  sin  6  cos  6  =  V  sin  2  e. 


M 


1.  A  carriage  runs  along  a  level  road  loith  velocttg  V,  throioing  off  particles 
of  mud  from  the  rims  of  its  wheels.  Find  the  greatest  height  to  which  any  of  them 
will  rise. 

Let  a  be  the  radius  of  the  wheel,  then  we 
have  seen  (p.  9)  that  any  point,  such  as  Q, 
moves  with  a  velocity  V  ■  QL/a  in  the  direc- 
tion QM  at  right  angles  to  QL.  This  will  be 
the  velocity  of  mud  projected  from  Q. 

If  the  angle  QLP  is  d,  the  height  above  the 
ground  at  which  the  mud  starts  is 


X              -^ 

■^^ 

^^o 

?\ 

1             ^ 

I 

/      \ 

J 

L 
Fig.  124 


The  mud  projected  with  this  vertical  velocity  will  attain  a  further  vertical 

height 

(Fsin2  6»)2 


29 


so  that  the  total  height  attained  is 


V2 


a  +  acos2e  -\ sin2  2  0. 

2^ 


This  may  be  written  as  a  quadratic  function  of  cos  2  0  in  the  foiTQ 


(-S 


F2\      T"2 

\a  -\ I cos2 20  +  a  cos 2 1 

"^9/       2gr 


/        F2\       a2gr       F2r       ^^      agT 

-la  +  — )  +  —^ cos  2  (9 ^ 

V        2gJ      2F2      2gl  F^J 


The  maximum  value  of  this  expression,  as  0  varies,  occurs  when  cos  2  0  =-^,, 

if  it  is  possible  for  cos  2  ^  to  have  this  value  —  i.e.  if  V-  >  ag.    In  this  case  the 
maximum  height  attained  is 

2g      2V^  2gV^ 

measured  from  the  ground. 

If,  however,  F2  <  ag,  we  cannot  make  cos  2  0 ~^  \  vanish.  We  accord- 
ingly make  it  as  small  as  possible,  so  that  we  take  cos  2  ^  =  1.  Thus  the  mud 
which  carries  to  the  highest  point  is  that  which  starts  at  the  top  point  M  of  the 
wheel,  and  obviously  this  never  gets  higher  than  its  starting  point. 


214 


MOTION  UNElER  CONSTANT  FOECES 


2.  Find  what  area  of  a  vertical  wall  can  be  covered  by  a  fire-hose  projecting 
water  with  velocity  v  at  a  distance  hfrom  the  wall. 

Let  S  be  the  nozzle  of  the  fire-hose,  and  let 
us  regard  it  as  capable  of  projecting  particles 
of  water  in  any  direction  we  please  with  a 
velocity  v.  The  points  which  can  be  reached 
will,  by  §  172,  be  all  the  points  which  lie  inside 
a  paraboloid  of  revolution  having  its  axis  ver- 

tical,  S  for  focus,  and  latus  rectum If  we 

9 
take  S  as  origin,  and  the  vertical  through  S 
for  axis  of  z,  the  equation  of  this  paraboloid 
will  be 


(a;2  +  2/2) 


Fig.  125 


2  V2  /  u2  \ 


The  curv^e  in  which  this  cuts  the  vertical  wall,  of  which  the  equation  may 
be  taken  tohe  y  =  h,  will  be 


x2  +  h"^ 


or 


9    \^9        I 

^,^2v^(v^_h^_\ 
9    \^9      2t32        / 


having  its  axis  vertical,  and  its 


2v- 
This  is  a  parabola,  of  latus  rectum  — , 

vertex  at  a  height  ^ 

V2    ^  k^g 

2g      2«2 

above  S.  All  the  points  inside  this  parabola  will  be  within  range  of  the  jet  of 
water.  The  points  on  the  wall  which  are  outside  this  parabola  will  be 
inaccessible. 


EXAMPLES 

1.  A  revolver  is  fired  horizontally  from  the  top  of  a  tower  100  feet  high,  the 
bullet  leaving  the  muzzle  with  a  velocity  of  600  feet  per  second.  Where  will  the 
bullet  strike  the  ground  ? 

2.  A  rifle  bullet,  fired  horizontally  at  a  height  of  10  feet  above  the  surface  of 
a  lake,  .strikes  the  water  at  a  distance  of  500  yards.  Find  its  velocity  in  feet 
per  .second,  the  resistance  of  the  air  being  supposed  negligible. 

3.  Prove  that  the  claim  for  a  rifle,  that  the  bullet  does  not  rise  more  than 
one  inch  in  a  range  of  100  yards,  implies  that  the  velocity  must  be  greater  than 
2078  feet  per  second. 

4.  Find  the  greatest  range  on  a  horizontal  plane  of  a  cricket  ball  thrown 
with  a  velocity  of  100  feet  per  second. 


EXAMPLES  215 

5.  A  shot  fired  from  a  gun  whose  muzzle  is  close  to  the  ground  just  clears  a 
man  6  feet  high  standing  10  yards  away,  and  embeds  itself  in  the  ground  a 
quarter  of  a  mile  off.  Show  that  the  shot  rises  to  a  height  above  the  ground 
which  is  certainly  greater  than  22  yards. 

6.  A  projectile  has  a  maximum  horizontal  range  of  2r)6  feet ;  what  is  its 
Telocity  of  projection  ? 

If  it  be  projected  with  this  velocity  from  a  point  on  the  floor  of  a  long 
corridor  24  feet  high,  what  will  be  its  greatest  range,  if  it  is  not  to  strike  the 
ceiling  ? 

7.  Prove  that  the  velocity  required  for  a  range  of  20  miles  will  be  at  least 
1840  feet  per  second,  with  a  time  of  flight  of  81.3  seconds. 

8.  Determine  the  charge  of  powder  required  for  the  range  of  20  miles  in  the 
last  question,  supposing  the  shot  to  weigh  a  ton,  and  the  strength  of  the  powder 
capable  of  realizing  100  foot-tons  per  pound  of  powder. 

9.  Show  that  the  range  i?  of  a  projectile  fired  from  a  height  h  above  a  level 
plane  with  velocity  v  at  an  angle  a  is  given  by 

2,v"-(h  +  R  tan  a)  =  gE^  sec^  a. 

10.  Show  that  the  area  of  a  level  plane  swept  by  a  gun  at  a  height  h  above 
the  plane  increases  proportionally  with  /;,  being  equal  to 

A  +  '2h  VttA, 

where  A  is  the  area  commanded  when  the  gun  is  at  the  level  of  the  plane. 

11.  A  projectile  can  be  fired  with  a  velocity  of  1720  feet  per  second  from  a 
fort  at  a  height  of  300  feet  above  a  horizontal  plane.  Find  what  area  of  the 
plane  is  covered  by  the  gun. 

12.  A  particle  is  projected  so  as  just  to  graze  the  four  upper  corners  of  a 
regular  hexagon  of  side  a,  placed  vertical  with  one  edge  resting  on  a  horizontal 
table  Find  the  highest  point  in  the  flight  of  the  particle,  and  show  that  the  range 
on  the  table  is  a  V?. 

13.  A  machine-gun  is  placed  on  an  armored  train  which  runs  along  a  hori- 
zontal line  of  rails  with  velocity  v.  The  muzzle  velocity  of  shots  fired  from  the 
gun  is  V.    Find  the  greatest  range 

(a)  in  front  of  the  train  ; 
(6)  behind  the  train. 

GENERAL  EXAMPLES 

1.  A  train  is  going  at  GO  miles  an  hour,  when  it  comes  to  a  curve  hav- 
ing a  radius  of  |  of  a  mile.  There  is  a  perfectly  smooth  horizontal  shelf 
in  the  train,  its  edge  being  parallel  to  the  rails  and  on  the  side  of  the  shelf 
away  from  tlie  center  of  the  curve.  A  small  object  stands  on  the  slielf  at 
a  distance  of  8  inches  from  the  edge.  Show  that  the  object  will  fall  off 
the  shelf  after  the  car  containing  the  object  has  described  about  24  yards 
of  the  curve,  and  find  what  its  horizontal  velocity  will  be  when  it  leaves 
the  shelf. 


216     MOTION  UNDER  CONSTANT  FOECES 

2.  A  balloon  is  moving  npwards  with  a  speed  which  is  increasing  at  the 
rate  of  4  feet  pei-  second  in  each  second.  Find  how  much  the  weight  of  a 
body  of  10  pounds,  as  tested  by  a  spring  balance  on  it,  would  differ  from 
its  weight  under  ordinary  circumstances. 

3.  An  Atwood's  machine  is  placed,  with  the  string  clamped,  on  one 
scale  of  a  weighing  machine.  Show  that  as  soon  as  the  string  is  undamped 
the  apparent  weight  of  the  machine  is  diminished  by 

(m  —  m')- 
III  +  w< 

where  m,  vi'  are  the  suspended  weights. 

4.  A  uniform  chain  of  length  I  and  weight  W  j)asses  over  a  smooth 
peg,  hanging  vertically  on  each  side.  If  the  chain  be  running  freely,  prove 
that  when  the  length  on  one  side  is  x,  the  pressure  on  the  peg  is 

4x£-^ 

5.  A  jet  of  water  falls  vertically  from  a  hose  to  the  ground,  starting  with 
a  velocity  which  is  negligible.  Show  that  the  center  of  gravity  of  the 
water  which  is  in  the  air  at  any  instant  is  two  thirds  of  the  way  up  from 
the  ground  to  the  hose. 

6.  A  heavy  uniform  chain  of  weight  w  is  tied  to  a  string  which  is  pulled 
up  with  tension  T.    Find  the  tension  in  the  chain  at  any  point. 

7.  Prove  that  the  shortest  time  from  rest  to  rest,  in  which  a  chain, 
which  can  bear  a  steady  load  of  P  tons,  can  lift  or  lower  a  weight  of  W 
tons  throuah  a  vertical  distance  of  h  feet  is 


V{^— 1  seconds. 


8.  In  a  system  of  pulleys  with  one  fixed  and  one  movable  block,  in 
which  the  cord  is  attached  to  the  axis  of  the  movable  block,  then  passes 
over  the  fixed  one,  then  under  the  movable  one,  and  then  over  the  fixed 
one,  find  the  weight  P  which,  when  attached  to  the  cord,  will  sujjport  a 
given  weight  W  hung  from  the  movable  block.  (The  blocks  are  so  small 
that  all  the  straight  portions  of  the  cord  may  be  considered  parallel.) 

Show  that,  if  the  weights  do  not  balance,  the  downward  acceleration  of 
W,  when  let  go,  will  be  tta  _  a  p 

W  -V  9P^' 

the  weight  of  the  cord  being  neglected,  and  that  of  the  movable  block 
being  included  in  W. 

9.  A  pulley  carrying  a  total  load  W  is  hung  in  a  loop  of  a  cord  which 
passes  over  two  fixed  pulleys,  and  has  weights  P  and  Q  freely  suspended 
from  its  ends,  each  segment  of  the  cord  being  vertical;  show  that,  when 


EXAMPLES  217 

the  system  is  let  go,  W  will  remain  at  rest  or  move  with  uniform  velocity, 

114 

provided 1 =  —  and  there  is  no  friction  anywhere. 

^  P      Q      W  ^ 

If  this 'relation  does  not  hold,  find  the  acceleration  of  \V. 

10.  A  particle,  falling  rmder  gravity,  describes  100  feet  in  a  certain 
second.  How  long  will  it  take  to  describe  the  next  100  feet,  the  resistance 
of  the  air  being  neglected  ? 

If,  owing  to  resistance,  it  takes  .9  second,  find  the  ratio  of  the  resist- 
ance (assumed  to  be  constant)  to  the  weight  of  the  particle. 

11.  Prove  that  the  line  of  quickest  descent  from  aiiy  curve  to  any  other 
curve  in  the  same  vertical  plane  makes  equal  angles  with  the  normals  to 
the  two  curves  at  the  points  at  which  it  meets  them. 

12.  Find  the  position  of  a  point  on  the  circumference  of  a  vertical 
circle,  in  order  that  the  time  of  rectilinear  descent  from  it  to  the  center 
may  be  the  same  as  that  to  the  lowest  point. 

13.  Find  the  line  of  quickest  descent  from  the  focus  to  a  parabola  of 
which  the  axis  is  vertical  and  vertex  upwards,  and  show  that  its  length  is 
equal  to  the  latiis  rectum. 

14.  An  ellipse  is  suspended  with  its  major  axis  vertical.  Find  the  diam- 
eter down  which  a  particle  can  fall  in  the  least  time.  What  is  the  least  value 
of  the  eccentricity  in  order  that  this  diameter  may  not  be  the  major  axis  ? 

15.  A  bullet  is  to  be  fired  at  a  vertical  target  so  as  to  hit  it  exactly  at 
right  angles.  If  v  is  the  velocity  of  the  bullet  and  a  the  distance  of  the 
target  from  the  point  of  firing,  show  that  the  angle  of  elevation  of  the 

shot  must  be  J  sin~i|  — -  ),   and  show  that  the  point  of  the  target  which  is 

hit  will  be  at  half  the  height  of  the  point  aimed  at. 

16.  A  bullet  is  fired  at  a  vertical  target.  Show  that  the  projection  of 
the  bullet  on  the  target,  as  seen  by  the  firer  of  the  shot,  appears  to  move 
with  uniform  velocity. 

17.  A  gun  fires  two  shots,  one  with  velocity  v  at  elevation  a,  and  the 
second  with  velocity  v'  at  a  smaller  elevation  a',  in  the  same  vertical  plane. 
Show  that  the  shots  will  collide  if  the  interval  between  firing  is 

2   vv'  sin(a  —  or') 
g  V  cos  a  +  v'  cos  a' 

18.  A,  B,  C  are  three  points  in  order  in  a  horizontal  line,  AB  being 
640  feet ;  a  particle  is  projected  from  A  with  a  velocity  of  390  feet  per 
second  in  a  direction  making  an  angle  tan~  ^j%  with  AC  ;  at  the  same  instant 
another  particle  is  projected  from  B  with  a  velocity  of  250  feet  per  sec- 
ond in  a  direction  making  an  angle  tan-^  J  with  BC ;  show  that  these  par- 
ticles will  collide,  and  find  when  and  where. 


218      MOTION  UNDER  CONSTANT  FORCES 

19.  A  howitzer  gun  has  a  muzzle  velocity  of  400  feet  per  second.  AVhat 
distance  immediately  behind  the  top  of  a  hill  200  feet  high  on  a  level  plane 
is  safe,  if  the  gun  is  distant  1000  yards  hxnn  the  point  in  the  plane  verti- 
cally below  the  top  of  the  hill  ? 

20.  The  sights  of  a  gun  are  inaccurately  marked,  the  gun  carrying 
always  ;}  per  cent  farther  than  the  distance  indicated  on  the  sights.  A 
nuirksman,  not  aware  of  the  error,  aims  at  a  target  distant  1000  yards.  If 
the  velocity  of  the  shot  is  1200  feet  per  second,  show  that  he  will  hit  the 
target  about  one  yard  too  high. 

21.  The  sighting  of  a  rifle  is  accurate,  and  to  aim  at  a  point  on  a  ver- 
tical target  distant  a  feet  the  rifle  ought  to  be  elevated  to  an  angle  a. 
Owing  to  the  unsteadiness  of  the  marksman's  hand,  the  rifle  is  pointed  in 
directions  which  lie  anywhere  within  a  small  angle  0  of  the  true  direction. 
Show  that  if  a  succession  of  shots  is  fired,  the  points  at  which  they  hit 
the  target  will  all  lie  within  a  snuiU  ellipse  of  semi-axes  ad  cos  a  and 
a0(l  —  tan^a)  respectively. 

When  a  =  —,  the  minor  axis  of  this  ellipse  vanishes,  so  that  the  shots 
4 
ought  all  to  lie  in  a  straight  line.     Interpret  this  result. 

22.  A  particle  slides  down  the  outer  surface  of  a  smooth  sphere,  start- 
ing from  rest  at  the  highest  point.  It  leaves  the  sphere  at  a  point  P  and 
describes  a  parabola  in  space.  Prove  that  the  circle  of  curvature  of  the 
parabola  at  P  will  touch  the  directrix. 

23.  A  particle  is  projected  horizontally  from  the  low^est  point  of  the 
interior  of  the  surface  of  a  smooth  sphere.  It  leaves  the  surface  of  the 
sphere  at  P,  and  after  describing  a  parabola  strikes  the  sphere  again  at  Q. 
Show  that  PQ  and  the  tangent  at  P  make  equal  angles  with  the  vertical. 

24.  Show  that  the  whole  area  commanded  by  a  gun  planted  on  a  hill- 
side, supposed  plane,  is  an  ellipse,  whose  focus  is  at  the  gun,  eccentricity 
the  sine  of  the  inclination  of  the  hill,  and  semi-latus  rectum  equal  to  twice 
the  height  to  which  the  muzzle  velocity  of  the  shot  is  due. 

25.  Show  that  the  area  of  a  plane  hillside  of  inclination  a,  which  is 
commanded  l)y  a  gun  placed  on  a  fort  of  height  //above  the  hill,  is 

4  rrh  (h  +  H  cos^  a)  sec^  a, 
where   v2  gh  is  the  muzzle  velocity  of  the  shot. 

26.  A  spherical  shell  of  mass  m  explodes  when  moving  with  negligible 
velocity  at  a  height  of  h  feet  above  the  ground.  The  shell  is  divided  into 
very  small  particles,  each  of  which  moves,  after  the  explosion,  away  from 
the  center  of  the  shell  with  velocity  v,  and  ultimately  falls  to  the  ground. 
Find  the  total  mass  of  the  fragments  which  will  be  found  per  unit  area  at 
any  specified  distance  from  the  point  vertically  underneath  the  shell. 


EXAMPLES  219 

27.  A  shell  bursts  while  in  the  air,  all  the  fragments  receiving  equal 
velocities  from  the  explosion.  Show  that  the  fragments  at  any  instant  lie 
on  a  sphere,  that  the  foci  of  the  paths  they  describe  also  lie  on  a  sphere, 
and  that  the  vertices  lie  on  a  spheroid. 

28.  A  particle  slides  down  a  rough  inclined  plane  AB,  starting  at  rest 
from  A  and  describing  a  parabola  freely  after  leaving  the  plane  at  B.    If 

i^is  the  focus  of  the  parabola  described,  show  that  the  angle  AFB  =  -  +  €, 
where  e  is  the  angle  of  friction. 

29.  From  a  fort  a  buoy  was  observed  at  a  dej'jression  i  below  the  horizon  ; 
a  gun  was  fired  at  it  at  an  elevation  a,  but  the  shot  was  observed  to  strike 
the  water  at  a  point  whose  depression  was  f .  Show  that,  in  order  to  strike 
the  buoy,  the  gun  must  be  fired  at  an  elevation  6,  where 

cos  9  sin  (d  +  /)   _  cos-  i  sin  i' 
cos  a  sin  (a  +  i')      cos"-^  i'  sin  i 

30.  Show  that  the  least  energy  which  will  project  a  jiarticle  over  a  wall 
which  is  at  distance  a  from  the  point  of  projection  is 

1  +  tan  1  a 

.7  nif/a = —  > 

1  —  tan  I  a 

where  a  is  the  elevation  of  the  top  of  the  wall  at  the  point  of  pi-ojection. 

31.  A  mill  wheel  of  radius  a  revolves  so  that  its  rim  has  a  velocity  V, 
and  drops  of  water  are  tlirown  oft'  from  the  rim  of  the  wheel.  Show  that 
the  envelope  of  their  paths  is  a  parabola  whose  axis  is  vertical  and  whose 

focus  is  at  a  distance  — —  vertically  above  the  center  of  the  wheel. 


CHAPTEE  IX 
MOTION  OF  SYSTEMS  OF  PARTICLES   " 

Equations  of  Motion 

176.  The  present  chapter  will  deal  with  the  motion  of  systems 
of  particles,  taking  account  of  the  actions  and  reactions  which  may 
be  set  up  between  the  different  pairs  of  particles.  As  a  prelimi- 
nary to  this,  it  will  be  convenient  to  recapitulate  the  results  which 
have  been  obtained  for  a  smgle  particle,  stating  these  results  in  a 
more  analytical  form  than  before. 

The  whole  system  of  forces  which  act  on  a  particle  must, 
since  they  act  at  a  point,  have  a  single  force  as  resultant.  Let 
us  call  this  resultant  P,  and  denote  its  components  along  three 
rectangular  axes  by  X,  Y,  Z. 

Also  the  particle,  being  regarded  as  a  point,  must  have  a  definite 
acceleration  /,  and,  since  /  is  a  vector,  this  acceleration  may  be 
supposed  to  be  compounded  of  three  components /^,/j,,/^  along  the 
three  coordinate  axes. 

The  second  law  of  motion  supplies  the  relation 

P  =  mf.  (65) 

We  are,  however,  told  more  than  tliis  by  the  second  law  of 
motion  :  we  are  told  that  the  directions  of  P  and  of  /  are  the  same. 
Let  \,  IX,  V  be  the  direction  cosines  of  this  single  direction,  then 
we  have  x  =  XP,      Y  =  fiP,     Z=  vP, 

and  also  f,  =  \f,      /,,  =  fu-f,      /,  =  vf. 

From  these  relations,  combined  with  relation  (65),  we  clearly 
have  X=mf^ 

Y=mfi  (66) 

Z=mfJ 

220 


EQUATIONS  OF  MOTION  221 

These  are  the  equations  of  motion  of  a  particle  in  analytical 
form.  They  simply  express  the  second  law  of  motion  in  mathe- 
matical language. 

177.  Let  X,  y,  z  be  the  coordinates  of  the  particle  at  any  instant, 
and  let  u,  v,  w  be  the  three  components  of  its  velocity.  The  com- 
ponent u  is  the  velocity,  along  the  axis  of  x,  of  the  projection  of 
the  moving  point  on  the  axis  of  x,  and  the  distance  of  this  point 
from  the  origin  at  any  instant  is  simply  x.  Thus,  by  the  defini- 
tion of  velocity,  we  have 

dx  ^g^^ 

and  similarly,  of  course, 


u 

dt 

V 

_dy 

dt 

dz 

to 

=  — . 

dt 

du 
The  rate  at  which  the  a;-component  of  velocity  increases  is  -—> 

but  it  has  also  been  supposed  to  be/,.,  for  tliis  is  the  ^-component 
of  the  acceleration.    Thus  we  have 


and  similarly 


Using  the  values  just  found  for  u,  v,  w,  these  equations  become 

.       d'x 
^^  =  ^' 
^d^ 
'''      df' 


/.= 

dii 
"di' 

/.= 

dv 
^~di' 

/.= 

dw 
dt 

J  ^         Jtl 


df 


222  MOTION  OF   SYSTEMS  OF  PARTICLES 

Substituting  these  expressions  for  the  components  of  accelera- 
tion into  the  equations  of  motion  (66),  we  obtain  these  equations  in 

the  new  form  ^2^>, 

X=  m-— 


df 
dH 


(68) 


178.  Suppose  we  have  a  system  of  particles,  —  m^  at  x^,y^,  z^; 
m^  at  «2J  2/2'  ^2  5  6tc.,  —  and  let  the  components  of  force  acting  on 
them  be  A\,  Y^,  Z^\  X^,  Y^,  Z^;  etc. 

Then,  from  the  equation  just  obtained, 

d^x 
X^  =  m^-^,  etc., 

so  that,  by  addition,  V  A'  =  V wi  — .  (69) 

where  ^  denotes  summation  over  all  the  particles  of  the  system. 
The  left-hand  member  ^X  is  the  sum  of  the  components  along 
Ox  of  all  the  forces  acting  on  all  the  particles  of  the  system.    As 
in  §  50,  these  forces  may  be  divided  into  two  classes : 

(a)  external  forces ' —  forces  applied  to  the  particles  from  outside 
the  system ; 

(h)  internal  forces  —  forces  of  interaction  between  pairs  of  par- 
ticles of  the  system. 

As  in  §  50,  we  find  that  the  contribution  to  2^X  from  the 
second  class  of  forces  is  nil.  For  all  these  forces  fall  into  pairs, 
each  pair  consisting  of  an  action  and  reaction,  of  which  the  com- 
ponents are  equal  and  opposite. 

Thus  in  calculating  ^X  we  need  only  take  account  of  exter- 
nal  forces. 


CONSERVATION   OF  LINEAR  MOMENTUM         223 

The  right-haud  term  of  ecjuation  (69),  namely  V//t  -7^'  can  also  be 

(JjX'  cJ^  r 

modified.    Since,  by  equation  (67),  we  have  %l  =  -^■t  the  value  of  — 

.    d.u         ,  «^^  ^^' 

IS  — )  so  that  79  7  7 

dt  NT^     ^  •*'     ^si:^     d\i       ft/vr^        \  ,^^, 

2:»;,,.=2;'»^=jX2;'»«)-        (^o) 

By  the  momentum  of  a  particle,  as  we  have  already  seen  (§  20), 
is  meant  the  product  of  its  mass  and  velocity.  The  momentum  of 
a  particle  is  therefore  a  vector  of  components  mii,  mv,  mw,  and  mu 
may  be  spoken  of  as  the  x-component  of  the  momentum.  Each 
particle  of  the  system  will  have  momentum,  and  the  sum  of  the 
a?-components  will  be  (  Vtow  \  the  quantity  which  appears  on  the 
right  hand  of  equation  (70). 

We  may  now  replace  equation  (69)  by 

where  2j^  denotes  the  sum  of  the  a;-components  of  the  external 
forces,  and  Vv/im  is  the  sum  of  the  ^--components  of  momentum. 

Conservation  of  Linear  Momentum 
179.  When  there  are  no  external  forces  acting,  ^A'  =  0,  so  that 

i&'"')  =  «•  (-2) 

d 
Similarly  we  have        ~r(S"^'")  —  ^>  ij'^) 

These  equations  express  that  the  quantities 

^inu,     ^mr,      ^mw 

do  not  vary  with  the  time.    That  is  to  say,  the  components  of  the 
total   momentum    are    constant,    so    that    the    total   momentum. 


224  MOTION  OF  SYSTEMS  OF  PARTICLES 

regarded  as  a  vector,  is  constant.    This  is  known  as  the  principle 
of  the  conservation  of  momentum.    Stated  in  words  it  is  as  foUows : 

When  any  system  of  particles  moves  ivithout  being  acted  on  by 
external  forces,  the  total  momentum  of  the  system  remains  constant 
in  magnitude  and  direction. 


Motion  of  Center  of  Gravity  of  System 

180.  Let  us  now  return  to  the   consideration  of  the  general 

equations  (71), 

Let  X,  y,  z  denote  the  coordinates  of  the  center  of  gravity  of  the 
particles  of  the  system  at  any  instant,  and  let  the  components  of 
the  velocity  of  this  point  be  denoted  by  u,  v,  w.    Then  we  have 

u  =  — ,  etc.  (76) 


dt 
The  value  of  x  is,  by  equation  (8), 

_   X 


mx 
X  = 


X' 


^  _      dx       d  iX'''^^ 

Tlius  1^  =  -r  =  -ri  ^r 

dt        dt  \   V  ; 


Sdx 


^m  ^m 

so  that  if  M  is  the  total  mass,  Vm,  of  all  the  particles,  we  have 

"^mu  =  Mu. 


MOTION   OF  CENTER  OF  GRAVITY  225 

Equation  (75)  now  becomes 

dt 


■^  du 


and  similarly  we  have  the  equations 

dt 
du 
Ift 


X^  =  J/-^-  (-9) 


Eemembering  that 


du     dv     dw 
dt      dt      dt 


are  the  components  of  acceleration  of  the  center  of  gravity,  we  see 
that  the  motion  of  the  center  of  gravity  is  the  same  as  it  would 
be  if  it  were  replaced  by  a  particle  of  mass  M,  acted  upon  by  a 

force  of  components  ^A',  ^Y,  2^Z.  This  force  again  is  simply 
the  force  which  would  be  the  resultant  of  all  the  external  forces, 
if  they  were  all  applied  to  the  imaginary  particle  which  we  are 
supposing  to  move  with  the  center  of  gravity. 

181.  In  the  particular  case  in  which  there  are  no  external  forces, 
the  center  of  gravity  moves  as  if  it  were  a  particle  acted  on  by  no 
forces,  so  that  its  motion  will  be  a  motion  of  uniform  velocity  in  a 
straight  line. 

182.  The  motion  of  the  center  of  gravity  m  this  particular  case, 
and  in  the  more  general  case  in  which  external  forces  act,  may 
accordingly  be  supposed  to  be  governed  by  the  two  following  laws  : 

Law  I.  The  center'  of  gravity  of  every  system  of  particles  con- 
tinues in  a  state  of  rest,  or  of  uniform  motion  in  a  straight  line, 
except  in  so  far  as  the  action  of  external  forces  on  the  system  com- 
pels it  to  change  that  state. 

Law  IL  When  external  forces  act  on  the  system,  the  motion  of 
the  center  of  gravity  is  the  same  as  it  wotdd  he  if  all  the  masses  of 
the  particles  were  concentrated  in  a  single  particle  which  moved 
with  the  center  of  gravity,  and  all  the  external  forces  were  apjjlied 
to  this  particle. 


22G  MOTIOX   OF   SYSTEMS  OF  PARTICLES 

These  laws  may  be  regarded  as  the  extensions  of  Newton's  Laws 
I  and  II  to  the  motion  of  a  system  of  particles.  We  can  see  now 
why  it  is  often  legitimate  to  apply  Newton's  second  law  to  the 
motion  of  bodies  of  finite  size,  as  though  they  were  particles 
(cf.§26)._ 

The  principle  of  conservation  of  momentum  is  often  sufficient 
in  itself  to  supply  the  solution  of  a  dynamical  problem  in  which 
only  two  bodies  are  in  motion. 

ILLUSTRATIVE  EXAMPLE 

A  shot  of  mass  m  is  fired  from  a  gun  of  mass  M,  which  is  free  to  run  back  on 
a  pair  of  horizontal  rails.  Find  the  velocity  of  recoil  of  the  gun,  and  examine  the 
influence  of  the  recoil  on  the  motion  of  the  shot. 

Let  us  suppose  that,  before  firing,  the  gun  stands  pointing  at  an  angle  a  to  the 
horizon,  and  let  the  muzzle  velocity  of  the  shot  —  i.e.  the  velocity  relative  to 
the  gun  with  which  the  shot  emerges  — be  V. 

Let  us  suppose  that  the  velocity  of  the  shot  relative  to  the  earth  has  compo- 
nents u,  V  horizontal  and  vertical,  and  let  the  velocity  of  recoil  of  the  gun  be  U, 
measured  in  the  horizontal  direction  opposite  to  that  in  which  the  gun  is  pointing. 

The  system  consisting  of  the  gun,  powder,  and  shot  is  not  free  from  the 
action  of  external  forces,  but  these  forces,  namely  the  weight  of  the  system  and 
its  reaction  with  the  earth,  have  no  horizontal  component.  Thus  the  horizontal 
momentum  of  the  system  must  remain  unaltered  by  the  explosion.  This  hori- 
zontal momentum  was  zero  initially  :  it  is  therefore  zero  when  the  shot  leaves 
the  gun.    Thus  we  have,  neglecting  the  weight  of  the  powder, 

MU  -  mu  =  0.  (o) 

The  velocity  of  the  shot  relative  to  the  gun  has  components 

u  +  Z7,     V. 

This  velocity  must,  however,  be  a  velocity  V  making  an  angle  a  with  the 
horizontal.    We  therefore  have 

u  +  U  =  V  cos  a,  (6) 

V  =  V  sin  a.  (c) 

From  equations  (a)  and  (&)  we  find 

u  _  U  _V  cos  a 
M      m      M  +  m 

Thus  the  velocity  of  recoil  is 

U  = V  cos  a. 

M  +  m 


ILLU STEATITE  EXAMPLE  227 


The  components  of  actual  velocity  of  tlie  sliot  are 

T  cos  a, 


M  +  m 
V  =  V  sin  a. 


Thus  the  actual  velocity  of  the  shot  is 

Vm2  +  vi  =  V\l '^ — -  cos2a    , 

while  the  angle  of  elevation,  6,  is  given  by 


V      M  +  7n 

tan  0  =  ~  = tan  a. 

u  M 


EXAMPLES 


1.  An  empty  railway  truck  weighing  8  tons,  originally  at  rest,  is  run  into  by 
a  similar  truck  carrying  a  load  of  24  tons  and  moving  at  the  rate  of  a  mile  an 
hour,  and  the  two  trucks  then  move  on  together.    Find  their  common  velocity. 

2.  A  gun  of  mass  M  fires  a  shot  of  mass  m  horizontally.    Show  that  of  the 

work  done  by  the  powder,  a  fraction is  wasted  in  producing  the  recoil 

of  the  gun. 

3.  A  particle  of  mass  m  slides  down  a  smooth  inclined  plane  of  angle  a,  the 
plane  itself  (mass  M)  being  free  to  slide  on  a  smooth  table.  Find  the  acceleration 
of  the  particle  and  the  plane. 

4.  A  shell  is  observed  to  explode  when  at  the  highest  point  of  its  path.  It 
is  divided  into  two  equal  parts  of  which  one  is  seen  to  fall  vertically.  Prove 
that  the  other  will  describe  a  parabola  of  which  the  latus  rectum  will  be  four 
times  the  latus  rectum  of  the  original  parabola. 

5.  A  shot  of  I  ounce  weight  strikes  a  bird  of  weight  5  pounds  while  in  the 
air.  At  the  moment  of  striking,  the  shot  has  a  horizontal  velocity  of  1000  feet 
a  second  and  the  bird  is  flying  horizontally  in  the  same  direction  at  a  height  of 
64  feet  above  the  ground,  with  a  velocity  of  20  feet  a  second.  Show  that  the 
bird  will  fall  at  a  distance  of  about  52.2  feet  beyond  the  place  where  it  was 
struck  by  the  shot. 

6.  A  ship  of  5000  tons  steaming  at  20  knots  suddenly  nins  into  a  whale 
whose  weight  is  12  tons,  asleep  on  the  surface  of  the  water.  Bj^  how  much  is 
the  ship's  speed  reduced  ?    (Neglect  motion  of  water.) 

7.  A  mail  package  weighing  2  hundredweight  is  thrown  out  from  a  train 
going  at  60  miles  an  hour,  with  a  horizontal  velocity  relative  to  the  train  of  11 
feet  per  second  at  right  angles  to  the  track.  It  falls  into  a  handcart  of  weight 
3  hundredweight,  which  is  free  to  move  on  a  level  platform,  its  wheels  being 
set  so  that  its  motion  will  make  an  angle  of  30  degrees  with  the  track  of  the 
train.    With  what  velocity  will  the  cart  start  into  motion  ? 


228  MOTION  OF  SYSTEMS  OF  PARTICLES 

8.  A  mass  of  8  pounds  moving  north  at  a  speed  of  10  feet  per  second  is  struck 
by  a  mass  of  (3  pounds  moving  east  at  14  feet  per  second,  and  its  direotion  of 
motion  is  thereby  deflected  tlirough  30  degrees,  while  its  speed  is  increased  by 
1  foot  per  second  ;  show  that  the  velocity  of  the  other  is  diminished  by  7.3  feet 
per  second,  approximately,  and  find  its  new  direction  of  motion. 

9.  Two  ice  yachts,  each  of  mass  M,  stand  at  rest  on  perfectly  smooth  ice, 
with  their  keels  in  the  same  direction.  A  man  of  mass  7?i  jumps  from  the  first 
to  the  second,  and  then  immediately  back  again  on  to  the  first.  Show  that  the 
final  velocities  of  the  yachts  are  in  the  ratio  of  M  +  m  :  M. 


Kinetic  Energy 

183.  We  may  best  begin  the  study  of  the  kinetic  energy  of  a 
system  of  particles  by  drawing  attention  to  a  difficulty  wMch  has 
not  so  far  been  encountered  in  the  present  book.  This  difficulty 
will  be  best  illustrated  by.  a  particular  example. 

Suppose  that  a  ship  is  moving  through  the  water  with  a  velocity 
of  20  feet  per  second,  and  that  a  person  on  deck  throws  a  ball  of 
mass  ?n  forward  with  a  velocity  of  30  feet  per  second  relative  to 
the  ship.  If  the  person  were  fixed  in  space,  we  might  say  that  the 
work  he  cUd  was  equal  to  the  final  kinetic  energy  of  the  ball,  and 
was  therefore  ^m{30y,  or  450  m. 

On  board  ship,  however,  the  ball  originally  had  a  velocity  of  20 
and  the  thrower  increases  this  velocity  to  50.  The  change  in  the 
kuietic  energy  of  the  ball  is  accordingly 

^,rt(50)--^m(20f, 

or  1050  m.  If  this  represents  the  work  done  by  the  thrower,  then 
we  are  driven  to  suppose  that  it  would  be  more  than  twice  as 
hard  to  throw  the  ball  on  board  ship  as  on  laud.  This  would 
clearly  be  erroneous. 

184.  The  error  lies  in  this,  that  the  thrower  not  only  imparts  a 
velocity  to  the  ball  but  also  to  the  ship.  If  he  throws  the  ball 
forward  he  must,  from  the  principle  of  conservation  of  momentum, 
impart  a  backward  velocity  to  the  ship,  of  momentum  equal  and 
opposite  to  the  forward  momentum  of  the  ball.  The  total  work  per- 
formed is  equal  to  the  change  produced  in  the  total  kinetic  energy 
of  the  ship  and  tlie  ball. 


KINETIC  ENERGY  220 

Since,  by  the  third  law  of  motion,  no  force  can  act  singly,  it  fol- 
lows that,  in  every  case  of  calculation  of  work  from  kmetic  energy, 
it  will  be  necessary  to  consider  the  kinetic  energy  of  more  than  one 
body.  For  instance,  a  man  throwmg  a  ball  on  land  will  not  only 
jerk  the  ball  forward,  but  will  also  jerk  the  whole  earth  backward, 
and  the  energy  of  both  must  be  taken  into  account,  or  we  shall  get 
erroneous  results. 

185.  A  second  difficulty,  closely  connected  with  the  first,  suggests 
itself  at  once.  Suppose  we  have  a  ball  thrown  with  a  velocity  v 
along  the  deck  of  a  ship  moving  with  velocity  V.  We  have  seen 
that  we  must  not  suppose  the  kinetic  energ}'  of  the  ball  to  be 
i-  mv"',  but  is  it  any  more  legitimate  to  suppose  it  to  be  ^m{v  -\-  F)^? 
For  the  sea  in  which  the  ship  sails  will  have  a  further  velocity  V 
in  consequence  of  the  earth's  rotation,  so  that  the  energy  might 
equally  well  be  taken  to  be 

lm{v+V+V'f, 

and  so  we  might  go  on  indefinitely.  Knowmg  of  no  frame  of 
reference  which  is  absolutely  at  rest,  it  would  seem  to  be  impos- 
sible to  find  the  true  value  of  the  kinetic  energy.  Moreover, 
it  ought  to  be  noticed  that  the  expressions  for  the  kinetic  energy 
referred  to  different  frames  of  reference  differ  by  more  than  mere 
constants.  For  instance,  the  difference  between  the  two  expres- 
sions we  have  found  for  kmetic  energy  relative  to  the  sea  and 
kinetic  energy  relative  to  the  earth's  center  is 

1  m  (y  +  F-f  V'f  -  i  m  (v  +  Vy 

=  ^viV''+  mV'{i-  +V). 

This  difference  not  only  depends  on  m  and  V'  but  also  on  v  and  V. 
It  is  not  a  constant  difference,  and  so  does  not  disappear  when  we 
calculate  the  increase  in  kinetic  energy  resulting  from  the  action 
of  forces. 

The  theorems  which  follow  serve  the  purpose  of  showing  a  way 
through  these  and  similar  difficulties. 


230  MOTION  OF   SYSTEMS  OF  PARTICLES 

186.  Theorem.  The  kinetic  energy  of  any  system  of  moving  par- 
ticles is  equal  to  the  kinetic  energy  of  motion  relative  to  the  center 
of  gravity  of  the  particles,  plus  the  kinetic  energy  of  a  single  par- 
ticle of  mass  equal  to  the  aggregate  mass  of  the  system,  moving 
with  the  center  of  gravity. 

Let  the  particles  be  m^  at  a\,  y^,  z^,  etc.,  and  let  the  coordinates 
he  measured  with  the  center  of  gravity  taken  as  origin.  Let  the 
velocities  be  denoted  by  u.^^,  v^,  to^,  etc.,  and  let  these  also  be  meas- 
ured relative  to  a  frame  moving  with  the  center  of  gravity,  so  that 

dx. 
u,  =  --^,  etc. 
'       dt 

Let  the  velocity  of  the  center  of  gravity  referred  to  any  frame 
of  reference,  moving  or  fixed  (provided  only  that  the  directions  of 
the  axes  do  not  turn),  have  components  u,  v,  w.  Then  the  velocity 
of  the  particle  m^  is  compounded  of  the  velocity  of  the  particle,  rel- 
ative to  the  center  of  gravity  of  the  system,  of  components  u^,  v^, 
10^,  together  with  the  velocity  of  the  center  of  gravity,  of  com- 
ponents u,  V,  w.    Thus  the  whole  velocity  of  the  particle  m^  has 

components  _  _  _ 

tL  +  ti^,  v  +  v^,  w  -1-  w^. 

The  kinetic  energy  of  the  first  particle  is  accordingly 

i  m^  [(7^  +  n^''  +  {v  +  v^^  +  («^  +  10^% 

so  that  the  kinetic  energy  of  the  system  is 

1  ^vi\fu  +  uf  +  (^  +  vf  +  (^  -f-  wf\, 

or,  on  expanding  squares, 

\(^m^iji^  +  v'  +  w'') 

-\-  u'^mu  4-  v'^mv  +  w'^mw 

+  1  ^m(u^  +  7-2  +  iv%  (80) 

Since,  when  the  center  of  gravity  is  taken  as  origin,  the  coor- 
dinates of  the  particles  are  x^,y^,z^,  etc.,  we  have,  by  equations  (8), 


KINETIC  ENERGY  231 

0  =  ~ — >  etc., 

so  that  ^mx  =  0.     It   follows  that  Vm  -j-  =  0>    or  Vrnw  =?  0. 

Similarly  V??iv  =  0  and  Vm«'  =  0.  The  whole  of  the  second 
line  of  expression  (80)  is  now  seen  to  disappear,  so  that  we  find 
for  the  kinetic  energy  the  expression 

U^mYu^  +  v'  +  w')  +  ^^w^r  +  v^  +  w-),  (81) 

which  proves  the  theorem. 

187.  Next,  suppose  that  the  coordinates  of  the  center  of  gravity 
at  any  instant,  referred  to  an  imaginary  set  of  fixed  axes,  are 
X,  y,  z,  the  velocity  of  the  center  of  gravity  having,  as  before,  com- 
ponents U,  V,  w. 

We  have  supposed  that,  relative  to  the  center  of  gravity,  the 
particle  m^  has  coordinates  x^^,  y^,  z^,  and  components  of  velocity 
%,  v^,  Wy  Thus,  referred  to  the  imaginary  fixed  axes,  the  coordi- 
nates of  the  particle  mj  will  be 

x  +  x^,     y  -\-y^,     z  +  01, 

while  its  components  of  velocity,  as  before,  are 

U  +  U.^,      V  +  l\,      w  +  w^. 

Let  the  force  applied  to  the  particle  ^n^have  components  ^i,Yi,  Z\. 
As  in  §  141,  the  work  done  on  this  particle  by  the  external  forces 
is  equal  to  minus  the  work  performed  by  the  particle  against  these 
forces.  Thus  the  work  done  on  the  particle  while  it  moves  over 
any  small  element  of  its  path  is,  by  §  118,  equal  to 

X,d(x  +  X,)  +  J\d{y  +  y,)  +  Z,d(z  +  z,). 

The  total  work  done  on  all  the  particles  in  any  small  displace- 
ment is  therefore 

X[^\d{'c  +  ^i\)  +  J\d{T,  +  y,)  +  Z,d{z  +  z,)l 

and  this  can  be  separated  into  two  parts  as  follows: 


232  MOTION  OF  SYSTEMS  OF  PARTICLES 

The  first  part  may  be  taken  to  be 

Xx,dx+^\\dy  +^Z,dz,  (82) 

while  the  second  is 

2)  a;  dx^  +  2 1\  di/,  +X^i  d-x'  (83) 

By  equation  (77),  we  have 

where  M  is  the  total  mass  of  the  system,  expressing  that  tlie 
center  of  gi-avity  moves  as  though  it  were  a  particle  of  mass  M 
acted  on  by  a  force  of  components  ^AT^,  ^^^i,  ^^i-  It  is  at  once 
clear  that  expression  (82)  represents  the  work  done  in  the  motion 
of  this  imaginary  particle,  and  this  we  know  must  be  equal  to  the 
increase  in  its  kinetic  energy. 

The  total  work  done  is  the  sum  of  expressions  (82)  and  (83). 
This  total  work  is  equal  to  the  increase  m  the  total  kinetic  energy 
of  the  system  (by  §  140),  and  tliis  again  (by  §  186)  is  equal  to  the 
increase  in  the  kinetic  energy  of  motion  relative  to  the  center  of 
gravity  of  the  particles,  plus  the  increase  in  the  kinetic  energy  of 
the  imaginary  particle  of  mass  M  moving  with  the  center  of  gravity. 

This  latter  increase,  as  we  have  just  seen,  is  represented  by 
expression  (82),  so  that  the  former  must  be  represented  by 
expression  (83). 

Thus  the  increase  in  kinetic  energ}"  relative  to  the  center  of 
gravity  is 

^{X^dx^  +  Y,di/,  +  Z,dz,), 

and  is  therefore  equal  to  the  work  done  by  the  forces,  calculated 
as  thorigh  the  center  of  gravity  were  at  rest. 

188.  Thus  we  see  that,  in  the  theorem  that  the  increase  in 
kmetic  energy  is  equal  to  the  work  done,  it  is  legitimate  to  calcu- 
late both  the  kinetic  energ}^  and  the  work  done  by  consideruig 
motion  relative  to  the  center  of  gravity  only ;  i.e.  the  system  may 
be  treated  as  though  its  center  of  gra\ity  remained  at  rest. 


IMPULSIVE  FORCES  233 

As  an  illustration,  consider  the  problem  of  firing  a  shot  on  board  a 
moving  ship.  The  mass  of  the  shot  being  small  compared  with  that  of  the 
ship,  we  may  suppose  the  center  of  gravity  of  shot  and  shi[)  to  have 
exactly  the  motion  of  the  ship.  The  velocity  of  the  shot  relative  to  this 
center  of  gTavity  may  accordingly  be  taken  simply  to  be  that  relative  to 
the  deck  of  the  ship.  The  work  done  by  the  powder  in  ejecting  the  shot 
from  the  barrel  is  the  same  as  though  the  ship  w^ere  at  rest,  so  that  the 
velocity  of  the  shot  relative  to  the  ship  will  be  the  same  as  though  the 
ship  were  at  rest. 

EXAMPLES 

1.  A  cart  is  moving  with  velocity  V  and  a  man  on  the  cart  throws  out  sand 
horizontally  from  the  back  of  the  cart  at  the  rate  of  m  pounds  per  minute,  the 
sand  having  a  velocity  v  relative  to  the  road.    At  what  rate  is  the  man  working  ? 

2.  A  gun  capable  of  firing  a  shot  vertically  upwards  to  a  height  h  is  placed 
on  an  armored  train  running  with  velocity  V.  What  is  the  greatest  range  to 
which  a  sliot  can  reach  (a)  behind  the  train,  (6)  in  front  of  the  train  ? 

3.  In  the  last  question  find  the  nearest  point  to  tlie  track,  which  is  out  of 
range  of  the  gun. 

4.  A  shell  of  ma.ss  M  is  moving  with  velocity  V.  An  internal  explosion  gen- 
erates an  amount  E  of  energy,  and  thereby  breaks  the  sliell  into  masses  of  which 
one  is  k  times  as  great  as  the  other.  Show  that  if  the  fragments  continue  to 
move  in  the  original  line  of  motion  of  the  shell,  their  velocities  will  be 


V  +  V2  kE/M,         V  -  V2  E/kM. 

5.  Two  men,  each  of  mass  M,  stand  on  two  inelastic  platforms  each  of  mass 
m,  hanging  over  a  smooth  pulley.  One  of  the  men,  leaping  from  the  ground, 
could  raise  his  center  of  gravity  through  a  height  h.  Show  that  if  he  leaps  with 
the  same  energy  from  tlie  platform,  his  center  of  gravity  will  rise  a  height 

h(i 'I—Y 

\        2  (M  +  m)J 

Impulsive  Forces 

189.  There  are  many  instances  in  dynamical  problems  in  which 
the  action  of  a  force  begins  and  terminates  within  so  short  an 
interval  of  time  that  the  action  may  be  regarded  as  instantaneous. 
Such  forces  are  called  impulsive  forces.  As  instances  of  impulsive 
forces  we  may  take  the  forces  brought  uito  play  by  the  jerking  of 
an  inextensil)le  string,  or  by  the  collision  between  two  hard  bodies. 

The  change  of  momentum  produced  by  the  action  of  an  impul- 
sive force  is,  in  general,  of  finite  amount.    As  the  force  only  acts  for 


234  MOTION   OF   SYSTEMS  OF  PARTICLES 

an  iiifinitesimal  time,  the  rate  of  change  of  momentum  must  be  in- 
tiuite.  By  the  second  law  of  motion,  the  rate  of  change  of  momen- 
tum is  equal  to  the  force,  so  that  the  force  itself,  while  it  lasts, 
must  be  infinite.  Thus  an  impulsive  force  may  be  regarded  as  an 
infinite  force  acting  for  an  infinitesimal  time. 

190.  At  the  outset  of  our  study  of  impulsive  forces,  it  will  be 
well  to  notice  one  physical  peculiarity  of  these  forces.  A  perfectly 
ri<nd  body  was  defined  as  one  which  kept  its  shape  under  the 
action  of  any  forces,  no  matter  how  great.  At  the  same  time  it 
was  mentioned  that  no  perfectly  rigid  bodies  exist  in  nature. 
Under  the  action  of  infinite,  or  very  great,  forces  such  as  occur  in 
impulses,  no  body  may  be  treated  as  perfectly  rigid. 

The  consequence  of  this  is  that  when  any  impulsive  forces  are 
brought  into  action,  relative  motion  is  set  up  between  the  different 
small  particles  of  which  contmuous  bodies  are  composed.  This 
relative  motion  possesses  energy  of  a  kind  wliich  cannot  be  regained 
from  the  system  by  mechanical  processes ;  in  fact,  the  relative 
motion  of  these  particles  simply  represents  the  heat  of  the  body. 
Inasmuch  as  this  energy  cannot  be  recovered  from  the  system  as 
mechanical  work,  we  see  that  the  impulsive  forces  which  do  work 
in  producing  tliis  energy  cannot  be  treated  as  conservative  forces. 
Thus  we  see  that 

The  sum  of  the  potential  and  kinetic  energies  of  a  system  does 
not  remain  constant  through  the  action  of  impulsive  forces. 

For  clearly  part  of  the  total  energy  is  left,  after  the  impulses,  in  the 
form  of  heat. 

Consider,  for  instance,  a  lead  bnllet  striking  a  steel  target.  Supjiose 
that,  before  striking  the  target,  the  bnllet  is  moving  horizontally  with 
A^elocity  v  at  a  height  h.  Its  kinetic  energy  is  \mv^,  its  potential  energy 
being  mgli.  After  striking,  we  may  suppose  the  bullet  to  have  no  horizontal 
velocity,  but  to  fall  to  the  bottom  of  the  target.  At  the  instant  at  which 
this  fall  begins,  the  kinetic  energy  is  nil,  while  the  potential  energy  is  mgh, 
as  it  was  before  the  impact.  Thus  an  amount  of  energy  \mv^  has  disap- 
peared from  the  total  energy.  This  has  been  used  up  in  producing  motions 
of  the  particles  of  the  bullet  and  target  relative  to  one  another  ;  these  show 
themselves  in  the  form  of  heat,  and  also,  perhaps,  partly  in  permanent 
changes  of  shape,  —  a  dent  in  the  target,  or  a  flattening  of  the  bullet. 


IMPULSIVE  FORCES  235 

Measure  of  an  Impulse 

191.  The  change  of  momentum  produced  by  an  impulsive  force 
is  called  the  impulse  of  the  force.  Thus  if  an  impulse  /  acts  on 
a  mass  m,  changing  its  velocity  (or  component  of  velocity  in  tlie 
direction  of  the  impulse)  from  u  to  v,  we  have 

I=m{v  —  u).  (84) 

The  force  acting  at  any  instant  is,  by  the  second  law,  equal 
to  the  rate  of  change  of  momentum  of  the  particle  (or  body)  on 
which  it  acts.  If  the  force  is  of  constant  amount,  the  whole 
change  of  momentum  is  equal  to  the  product  of  the  force  by  the 
time  over  which  it  acts.  If  the  force  is  of  variable  amount,  the 
change  of  momentum  will  be  equal  to  the  integral  of  the  force 
with  respect  to  the  time  over  which  it  acts.  Thus  if  F  is  the 
value  of  the  force  acting  at  any  instant  of  the  whole  time  t,  we 
see  that  the  impulse 

=  Pf,  if  the  force  is  of  constant  amount, 


■i 


Fdt,  if  the  force  is  of  variable  amomit. 


Work  done  hy  an  Impulse 

192.  The  work  done  by  an  impulse  I  in  changing  the  velocity 
of  a  mass  m  from  ti  to  v  will  be 

\  mv^  —  i  mu"^, 

the  increase  in  the  kinetic  energy  of  the  mass.    Since  /=  m  {v  —  u), 
we  may  write  the  expression  for  the  work  done  in  the  form 

^m{v  —  u)  (r  +  u) 
_     ,'r  +  u'' 


9 

\       ^ 


Thus  the  work  done  hy  an  imptdse  is  equal  to  the  impulse  multi- 
plied hy  the  mean  of  the  initial  and  final  velocities  of  the  mass 
acted  upon. 


23G  MOTION   OF   SYSTEMS  OF  PAKTICLES 

If  the  mass  is  not  moving  in  the  same  direction  as  the  line  of 
action  of  the  impulse,  the  foregomg  result  will  obviously  be  true 
if  u,  V  are  taken  to  be  the  components  of  the  velocities  along  the 
line  of  action. 

ILLUSTRATIVE  EXAMPLES 

1.  A  shot  of  14  pounds  is  fired  into  a  tnrget  of  mass  200  pounds  which  is 
suspended  by  chains  so  that  it  is  free  to  start  into  motion  horizontally.  If  the 
shot,  before  impact,  ivas  moving  with  a  horizontal  velocity  of  1000  feet  a  second, 
and  afterwards  remains  embedded  in  the  target,  find  the  loss  of  energy  caused 
by  the  impact. 

Let  V  denote  the  liorizontal  velocity,  measured  in  feet  per  second,  with 

which  the  target  and  bullet  together  start  into  motion  after  the  impact.    Then, 

by  the  conservation  of  momentum,  equating  the  momentum  before  the  impact 

to  that  after, 

1000  X  14  =  F  X  214, 

so  that  "r=i|fl|a. 

The  kinetic  energy  before  impact  was  i  •  14  •  (1000)2  ;  that  afterwards  is 
•^  •  214  •  F2.    Thus  the  loss  of  energy  is 

i  (14,000,000  -  214  F-2)  =  6,540,000  foot  poundals,  approximately. 

2.  A  heavy  chain,  of  length  I  and  ynass  m  per  unit  length,  is  held  with  a  length 
h  hanging  over  the  edge  of  a  table,  and  the  remainder  coiled  up  at  the  extreme  edge 
of  the  table.    If  the  chain  is  set  free,  find  the  velocity  at  any  stage  of  the  motion. 

Suppose  that  at  any  stage  of  the  motion  a  length  x  is  hanging  vertically,  so 
that  a  length  i  —  x  is  coiled  up  on  the  table.  After  an  infinitesimal  time  dt  let 
X  be  supposed  to  have  increased  from  x  to  x  +  dx.  Then  if  i;  is  the  downward 
velocity  of  the  chain,  we  clearly  have 

dx 
V  —  — 
dt 

At  the  beginning  of  the  interval  dt,  the  downward  momentum  of  the  chain 
was  that  of  a  mass  mx  moving  with  a  velocity  v.  It  was  accordingly  mvx.  At 
the  end  of  the  interval,  the  momentum  Ls  that  of  a  mass  m  (x  +  dx)  moving  with 
a  velocity  which  may  be  denoted  by  {v  +  dv).    Thus  the  gain  in  momentum  is 

m  (x  +  dx)  {v  +  dv)  —  mxv, 

or,  neglecting  the  small  quantity  of  the  second  order  dv  dx,  the  gain  is 

m(xdv  +  vdx). 


ILLUSTRATIVE   EXAMPLES  237 

The  gain  of  momentum  per  unit  time  is,  however,  by  equation  (71),  equal 
to  the  total  force  acting,  and  this  is  mgx  at  tlie  beginning  of  the  interval  dt  and 
ifng{x  +  dx)  at  the  end.  Neglecting  the  small  quantity  of  the  second  order  dxdt, 
we  find  that  the  total  gain  of  momentum  in  the  interval  dt  must  be  mgxdt. 

Thus  we  have 

m  {xdv  +  V  dx)  =  mg  x  dt 

dx 
—  max  — , 

X) 

oi',  simplifying,  u  x V  v-  —  gx. 

dx 

To  integrate  this  equation,  we  multiply  by  2  x,  and  then  we  obtain 
v'hfl  =  I  gx^  +  a  constant. 

To  determine  the  constant,  we  note  that  v  =  0  when  x  —  k,  so  that  the  value 
of  the  constant  must  be  —  |  gh^.    Thus  we  have 

'^      x^ 

giving  the  velocity  when  a  length  x  is  off  the  table.    When  the  last  particle  of 
the  chain  is  pulled  off,  the  value  of  x  is  I,  so  that  at  this  instant 

We  notice  that  this  value  of  v-  is  not  the  value  which  would  be  obtained  from 
the  equation  of  energy.  Clearly  this  equation  must  not  be  employed,  since 
impulses  are  in  action  all  the  time,  jerking  new  particles  of  the  chain  into 
motion. 


EXAMPLES 

1.  An  empty  car  of  10  tons  weight  runs  into  a  similar  car  loaded  with  .50 
tons  of  coal,  and  the  two  run  on  together  with  a  velocity  of  5  feet  per  second. 
What  was  the  velocity  of  the  first  car  originally,  and  what  was  the  amount  of 
the  impulse  between  the  cars  ? 

2.  A  stone  of  weight  \  ounce  is  dropped  on  to  soft  ground  from  a  height  of 
6  feet.    Find  the  impulse  exerted  before  the  stone  is  brought  to  rest. 

3.  A  mass  of  1  ton  falls  from  a  height  of  10  feet  on  the  end  of  a  vertical 
pile,  and  drives  it  half  an  inch  deeper  into  the  ground.  Assuming  the  driving 
force  of  the  mass  on  the  pile  to  be  constant  while  it  lasts,  find  its  amount  and 
the  duration  of  its  action. 

4.  A  body  of  mass  10  grammes  is  moving  with  a  speed  of  8  centimeters 
a  second.  Suddenly  it  receives  a  blow  which  causes  it  to  double  its  speed,  and 
to  change  its  direction  through  half  a  right  angle.  Determine  the  direction  of 
the  blow,  and  the  velocity  with  which  the  body  would  have  moved  off,  had  it 
been  at  rest. 


238  MOTION  OF   SYSTEMS   OF  PAllTICLES 

5.  The  string  of  an  Atvvood's  machine  has  masses  mi,  m^  attached  to  its 
ends,  Ml  being  tlie  heavier.  After  it  has  been  in  motion  for  1  second  mi  strikes 
the  floor.  Find  (a)  for  how  long  mo  will  continue  to  ascend,  (b)  witli  what 
velocity  vii  will  start  into  motion  again  when  the  string  becomes  tight. 

6.  On  a  certain  day  one  inch  of  rain  fell  in  10  hovirs,  the  drops  falling  with 
a  velocity  of  20  feet  per  second.  Find  the  average  pressure  per  stiuare  foot  on 
the  canvas  roof  of  a  tent,  supposed  horizontal,  produced  by  the  impact  of  the 
raindrops.    (One  cubic  foot  of  water  weighs  62 J-  pounds.) 

7.  The  earth,  moving  in  its  orbit  with  velocity  F,  runs  into  a  swarm  of  small 
meteorites,  of  density  one  kilogramme  to  the  cubic  mile,  moving  with  a  velocity 
V  in  a  direction  exactly  opposite  to  that  of  the  earth.  Find  the  rate  of  decrease 
of  the  earth's  .speed  in  con.sequence  of  its  bombardment  by  the  meteorites, 
and  find  also  the  increase  in  the  height  of  the  barometer  at  different  points  on 
the  earth's  surface, it  being  assumed  that  all  the  meteorites  are  dissipated  into 
dust  before  they  reach  the  ground.  (The  earth's  mass  is  0  x  10-^  grammes,  its 
diameter  is  7927  miles.) 

8.  A  uniform  chain  is  coiled  in  a  heap  on  a  horizontal  plane,  and  a  man 
takes  hold  of  one  end  and  raises  it  uniformly  with  a  velocity  v.  Show  that 
when  his  hand  is  at  a  height  x  from  the  plane,  the  pressure  on  his  hand  is  equal 

to  the  weight  of  a  length  x  -\ of  the  chain. 

^9 


Elasticity 

193.  It  is  a  matter  of  common  experience  that  if  we  drop  a 
ball  of  steel  on  to  a  hard  floor  it  rebounds  to  a  considerable  height, 
wliile  a  ball  of  wood  will  rebound  to  a  much  smaller  height,  and  a 
ball  of  wool,  paper,  or  putty  will  hardly  rebound  at  all. 

When  the  contact  between  the  surfaces  of  two  bodies  is  of  such 
a  nature  that  they  do  not  rebound  at  all  after  impact,  it  is  said  to 
be  perfectly  inelastic,  wliile  if  the  bodies  rebound,  the  contact  is 
said  to  be  elastic.    Obviously  there  are  varying  degrees  of  elasticity. 

Moment  of  Greatest   Compression 

194.  Proljably  the  collision  of  two  billiard  balls  supplies  the 
most  familiar  instance  of  an  impact  with  a  high  degree  of  elasticity. 
AVe  shall  discuss  this  impact  best  by  referring  the  motion  of  the 
second  ball  to  a  frame  of  reference  moving  with  the  first.  Before 
impact  the  center  of  the  second  ball  is  approaching  that  of  the  first. 


ELASTICITY  239 

after  impact  it  is  receding.  Hence  at  some  instant  during  the 
impact,  its  motion  must  have  changed  from  one  of  approaching  to 
one  of  receding ;  at  this  instant  the  distance  between  the  two 
centers  was  a  minimum. 

Suppose  that,  before  the  experiment,  we  had  chalked  the  two 
faces  of  the  balls  on  which  the  collision  takes  place.  On  examin- 
ing the  balls  after  impact  it  will  be  found  that  the  chalk  has  been 
disturbed,  not  only  at  a  single  point,  but  all  over  a  circle  of  con- 
siderable size,  —  perhaps  of  diameter  half  an  inch  for  billiard  balls 
moving  with  a  fair  velocity.  This  shows  that  at  the  moment  at 
wliich  the  centers  of  the  balls  w^ere  closest  to  one  another,  their 
distance  was  less  than  if  they  had  been  placed  in  contact  and  at 
rest,  —  the  balls  were  compressed. 

The  instant  at  which  the  two  centers  were  nearest  is  called  the 
onoment  of  greatest  compression. 

In  general,  for  any  two  surfaces  in  collision,  the  instant  at  which 
the  relative  velocity  along  the  common  normal  vanishes  is  called 
the  moment  of  greatest  compression.  Obviously  this  is  tlie  instant 
at  which  the  motion  of  the  two  surfaces  changes  from  one  of 
approach  to  one  of  recession. 

195.  By  the  time  the  moment  of  greatest  compression  is  reached, 
the  velocities  of  both  bodies  will,  in  general,  have  been  chanoed, 
so  that  forces  must  have  been  at  work  to  produce  this  change.  The 
wdiole  time  of  action  of  these  forces,  the  time  from  the  instant  at 
which  the  bodies  first  touch  to  the  instant  of  greatest  compression, 
is  so  small  that  these  forces  may  be  treated  as  impulsive.  The 
impulses  acting  on  the  two  bodies,  being  action  and  reaction,  must 
be  equal  and  opposite.  If  the  surfaces  are  smooth,  the  direction  of 
these  impulses  will  be  along  the  common  normal.  If  the  surfaces 
are  rough,  w^e  cannot  specify  the  direction  until  we  know  the  direc- 
tion of  sliding,  if  any,  of  the  surfaces  over  one  another.  In  either 
case,  let  us  denote  the  component  of  the  impulse  along  the  com- 
mon normal  by  I.  The  quantity  /  is  called  the  impulse  of  com- 
pression. Clearly  it  is  the  forces  of  which  tliis  impulse  is  composed 
which  reduce  the  relative  normal  velocity  to  zero. 


240  MOTION   OF   SYSTEMS   OF  PARTICLES 

After  the  moment  of  greatest  compression,  a  second  system  of 
forces  must  come  into  play  to  set  up  the  velocities  with  which  the 
bodies  separate  from  one  another.  In  fact,  at  the  instant  of  great- 
est compression,  the  compressed  parts  of  the  bodies  act  like  a  com- 
l)ressed  spring,  and  we  can  suppose  the  velocities  of  separation 
produced  by  the  action  of  this  imaginary  spring.  The  forces  which 
separate  the  bodies  may  again  be  treated  as  impulsive,  and  the 
component  of  this  impulse  along  the  common  normal  will  be 
denoted  by  /'.    The  impulse  /'  is  called  the  impulse  of  restitution. 

196.  When  the  motion  of  the  bodies  before  impact  is  known,  we 
can  calculate  the  velocities  at  the  instant  of  greatest  compression 
by  an  application  of  the  principle  of  conservation  of  momentum. 
It  is  therefore  possible  to  calculate  the  impulse  /,  the  impulse 
of  compression. 

The  amount  of  the  impulse  /',  on  the  other  hand,  depends  on 
the  nature  of  the  contact  between  the  two  bodies;  for  instance, 
if  the  bodies  are  perfectly  inelastic,  there  is  no  separation  at  all 
after  impact,  so  that  /'  =  0.  In  general,  it  is  found  as  a  matter 
of  experiment  that  the  impulse  /'  is  connected  with  the  impulse 
/  by  the  simple  law 

I'=€l, 

where  e  is  a  quantity  wliich  depends  only  on  the  nature  of  the 
contact  between  the  two  surfaces,  and  not  on  the  amount  of  the 
impulse  /.  The  quantity  e  is  called  the  coefficient  of  elasticity  for 
the  two  bodies. 

It  is  important  to  understand  that  this  coefficient  of  elasticity  is  a  quan- 
tity entii-ely  different  from  the  coefficients  or  elastic  constants  which  occur 
in  the  theory  of  elastic  solids.  Indeed,  the  term  coefficient  of  elasticitj/  is 
somewhat  unfoi-tunate  as  a  description  of  the  quantity  e  ;  what  is  measured 
is  resilience  rather  than  elasticity,  and  doubtless  coefficient  of  resilience  would 
be  a  better  description  than  coefficient  of  elasticity.  The  term  coefficient  of 
elasticity  has,  however,  been  generally  adopted. 

197.  The  value  of  e,  as  we  have  seen,  is  zero  for  perfectly  in- 
elastic bodies.  For  iron  impinging  on  lead,  the  value  of  e  is  about 
.14,  for  iron  on  iron  about  .66,  and  for  lead  on  lead  about  .20.    We 


PARTICLE  IMPINGING  ON  A  FIXED  SUKFACE     241 

notice  that  resilience  depends  on  the  nature  of  the  contact  between 
two  bodies,  being  in  this  respect  similar  to  the  coetlicient  of  fric- 
tion. The  resilience  does  not  arise  partly  from  one  body  and  partly 
from  the  other,  for  if  it  did  the  value  of  e  for  iron  impinging  on 
lead  would  be  mtermediate  between  the  values  fur  iron  on  iron 
and  for  lead  on  lead. 

As  examples  of  bodies  for  wliich  the  coefficient  of  elasticity  is 
large,  it  is  found  that  the  value  of  e  for  the  impact  of  two  ivory 
billiard  balls  is  about  .81,  while  for  glass  impinging  on  glass  it  is 
.94.  The  most  perfect  elasticity  (Conceivable  is  that  of  two  bodies 
for  which  g  =  1,  in  wdiicli  case  the  impulse  of  restitution  is  e(iual 
to  the  impvdse  of  compression.  The  bodies  in  this  case  are  spoken 
of  as  perfectly  elastic.  The  peculiarity  of  perfectly  elastic  bodies  is 
that  no  energy  is  lost  on  impact.  It  is  clear  that  the  value  of  e  can- 
not exceed  unity,  for  if  the  value  of  e  were  greater  than  unity,  tlie 
kinetic  energ}'  set  up  by  the  impulse  of  restitution  would  be  greater 
than  that  absorbed  by  the  impulse  of  compression,  so  that  the  total 
energy  would  be  increased. 

We  shall  now  apply  these  principles  to  some  important  cases 
of  impact. 

Pakticle  impinging  on  a  Fixed  Surface 

Direct  Im2)c(ct 

198.  Suppose  first  that  the  unpact  is  direct  —  ie.  that,  at  the 
instant  of  colhsion,  the  particle  is  moving  along  the  normal  to  the 
surface  at  the  point  at  which  it  strikes.  Let  7«  be  its  mass,  and  v 
its  velocity  before  impact.  At  the  moment  of  greatest  compression, 
the  particle  will  be  at  rest  relatively  to  the  plane,  so  that  its 
momentum  is  reduced  by  the  impulse  of  compression  from  mv 
to  0.    Thus  we  must  have 

/=  mv. 

If  e  is  the  coefficient  of  elasticity, 

■ 

/'  =  f  7  =  cmv. 


242 


MOTION   OF   SYSTEMS  OF  PARTICLES 


Thus  there  is  a  normal  impulse  of  amount  emv,  and  this  gener- 
ates a  velocity  ev  in  the  particle.  There  is  no  tangential  impulse, 
for  there  is  no  sliding  of  the  surfaces  past  one  another.  Thus  the 
velocity  of  rebound  is  a  velocity  ev  normal  to  the  surface. 

Oblique  Imjjcict :  Smooth  Contact 

199.  If  the  impact  is  oblique,  let  us  suppose  that  the  components 
of  velocity  along  the  tangent  plane  and  along  the  normal  before 
impact  are  u,  v.    As  before,  we  find  ~^_^  ^  I^ 


/=  mv, 


emv, 


Ol^  \i —  ^  M^  V 


so  that  the  normal  velocity  after  impact,  say  v',  is 


C  V-  =.  \~) 


If  the  contact  is  supposed  smooth,  there  can  be  no  force  in  the 
tangent  plane,  so  that  the  momentum  in  the  tangent  plane  remains 
unaltered.  Thus  the  velocity  in  the  tangent  plane  remains  equal 
to  w,  and  the  velocity  after  impact  will  be  one  of  components  u,  ev. 
Let  6  be  the  angle  which  the  velocity  before 
impact  makes  with  the  normal,  and  let  ^  be  the 
corresponding  angle  after  impact.    Then 

tan  6  =  -> 

V 

tan  (f)  =  —, 

ev 

so  that  tan  6  =  e  tan  (f). 

,^.  If  the  bodies  are  perfectly  elastic,  e  =  1,  so  that 

6  =  (f)]  i.e.  the  particle  rebounds  at  an  angle  equal 
to  the  angle  of  incidence.  Its  reflexion  obeys  the  same  law  as 
that  of  a  ray  of  light. 

If  the  bodies  are  imperfectly  elastic,  6  is  less  than  <p,  so  that 
the  path  is  bent  away  from  the  normal. 

TT 

If  the  bodies  are  perfectly  inelastic,  e  =  0,  so  that  <^  =  —  J    the 

particle  simply  slides  along  the  plane,  as  of  course   it  obviously 
must  since  /'  =  0. 


PARTICLE  lMPmGl]S"G  OX  A  FIXED  SUEFACE     243 
The  kinetic  energy  before  impact  is 

that  after  impact  is  %  m  [u"^  +  c'-). 

Thus  there  is  a  loss  of  kinetic  energy  of  amount 

or  ^7nv^{l  —  e^). 

This  vanishes  if  e  =  1,  i.e.  if  the  bodies  are  perfectly  elastic. 
In  all  other  cases  there  is  a  loss  of  energy.  We  again  see  that  e 
cannot  be  greater  than  unity,  or  it  would  be  possible  to  gain 
energy  by  causing  bodies  to  impinge  on  one  another. 

Oblique  Invpact :  Rough  Contact 

200.  As  in  the  case  of  a  smooth  contact,  we  obtain  the  relation 
v'  =  ev  connecting  the  components  of  velocity  along  the  normal. 
The  reaction,  however,  no  longer  acts  entirely  along  the  normal,  so 
that  it  is  not  now  true  that  the  tangential  component  of  velocity 
remams  unaltered. 

Let  us  consider  the  case  in  which  the  surface  of  the  particle 
slides  in  the  same  direction  over  the  fixed  surface  during  the 
whole  time  that  the  two  surfaces  are  in  contact.  Tlien  at  every 
instant  of  the  impact  tliere  will  be  a  tangential  force  equal  to 
/JL  times  the  normal  force,  so  that  the  total  tangential  impulse 
must  be  /m  times  the  total  normal  impulse,  and  therefore  equal 

to   fJi{I+l'). 

Thus  if  u'  is  the  tangential  velocity  after  the  impact,  we  have 

VI  {u—  u')  =  fx{I+  I') 
=  fi{l  +  e)I 
=  /A  ( 1  +  c)  mv, 
so  that  2('  =  n  —{1+  e)  fiv. 


244  M0T1()^■   OF  8Y8TEMS  OF  PARTICLES 

If,  as  before,  we  suppose  that  6,  (f)  are  the  angles  which  the  path 
makes  with  the  normal  before  and  after  the  impact  (see  tig.  126), 
we  have 


tan  e  = 

u 

'■  — ) 

V 

tan  cf>  = 

n'       u  ■ 

-(1  + 

ev 

e)fiv 

e  tan  (f)  = 

:  tan  ^  - 

a+^'] 

)fi. 

so  that 

The  value  of  {1  +  c)  /j,  is  always  positive,  so  that  (f>  is  less  than  it 
would  be  if  the  plane  were  smooth ;  in  other  words,  the  roughness 
of  the  plane  causes  the  particle  to  rebound  nearer  to  the  normal. 

This  equation,  however,  is  only  true  within  certain  limits,  for 
we  have  assumed  that  there  is  sliding  during  the  whole  time  of 
impact.  It  may  be  that  at  a  certain  stage  of  the  motion  sliding 
will  give  place  to  rollmg,  and  if  so  the  equation  we  have  obtained 
is  no  longer  valid. 

Impact  of  Two  Moving  Bodies 

201.  Suppose  now  that  two  bodies  A,  B  of  masses  m,  m'  im- 
pinge at  the  point  C,  the  common  normal  to  C  being  the  line  CP. 
Let  it  be  supposed  that  the  centers  of 
gravity  of  the  two  bodies  both  lie  in  the 
line  CP  at  the  moment  of  impact,  and  let 
the  components  along  CP  of  the  velocities 
of  the  centers  of  gravity  of  the  masses 
A,  B  respectively  be 

u,  li'  before  impact, 
V,  V  at  the  instant  of  greatest 
compression, 
Fig-  127  /ji^  ^'  after  impact. 

Then  if  we  denote  the  impulse  of  compression  by  /,  and  the 
impulse  of  restitution  by  /',  we  have 

I  =  ni  (u-V)=-  m'  (u'  -  V),  (85) 

/'  =  m  {V-  v)  =  -  m' {V-  v').  (86) 

-  --  ?-        .  \A- 


IMPACT  OF  TWO  MOVING  BODIES  245 

From  the  first  Hue  u  =V-{ > 

ill 

III 

so  that  10—  u'  =  I { 

\m      111' 

an  equation  connecting  /  with  the  rehitive  velocity  before  collision. 
Similarly,  from  equations  (86), 

The  experimental  relation  I'  —  el  is  now  seen  to  be  exactly 
equivalent  to  the  relation 

v  —  v'=-  —  c(u  —  u'),  (87) 

or,  in  words :  The  noi^mal  component  of  relative  velocity  of  the 
centers' of  gravity  after  collision  is  equal  to  e  times  the  relative 
velocity  before  collision,  and  is  in  tlie  o-ppositc  direction. 

This  law  is  known  as  Newton's  experimental  law ;  it  expresses 
the  same  property  of  matter  as  the  relation  I'  —  el. 

A  second  relation,   connecting  velocities    before    impact    with 

velocities  after,  is  given  by  the  conservation  of  momentum ;  we 

have  ,      /  r  ,      I  r 

mv  +  m  V  =  mu  +  m  ii . 

Combinmg  this  with  equation  (87),  we  can  determine  the 
velocities  v,  v'  after  collision  in  terms  of  the  velocities  u,  n' 
before  collision. 

Solving,  we  find  that 

_  nut  +  m'u'  —  em'{u  —  v.') 


V  = 


m  -\-  m' 
mil  +  m'li'  +  cm  (u 


(88) 
(89) 


m  +  III' 
giving  the  normal  velocities. 

If  the  bodies  are  rough,  we  find  the  tangential  velocities  in 
the  same  way  as  in  §  200;  while  if  the  bodies  are  smooth,  the 
velocities  in  directions  perpendicular  to  CP  remain  unaltered. 


246  .MOTION   OF   SYSTEMS  OF  PARTICLES 

If  the  center  of  gra\dty  of  the  two  bodies  is  at  rest,  —  or,  what 
comes  to  the  same  thing,  if  we  measure  all  velocities  relatively  to 
the  center  of  gravity,  —  we  have 

nut  +  tii'u'  =0, 

m'(u  —  u') 
so  that  V  =  —  e 


m  +  7n' 


m  (u  —  u') 


m  +  111' 
Using  the  relation  mu  =  —  m\i' ,  these  become 

V  ■=  —  eu, 
v'  =  —  eu' , 

so  that  the  bodies  rebound  from  one  another  as  though  they  had 
impinged  on  a  fixed  plane  of  elasticity  e. 

The  kinetic  energy,  either  before  or  after  the  collision,  is  equal 
to  the  kinetic  energy  of  a  single  particle  movmg  with  the  center  of 
gravity,  together  with  that  of  the  system  relative  to  the  center  of 
gravity.  The  former  remains  unchanged  by  collision,  so  that  the 
loss  in  the  total  kinetic  energy  produced  by  collision  is  equal  to 
the  loss  in  the  kinetic  energy  relative  to  the  center  of  gravity. 

If  the  bodies  are  smooth,  this  loss  of  kinetic  energy 

=  ^{m?<.^+  m'u''—  mr^—  m'v'^) 
=  ^(mn'^-\-  m'ti'')  (I—  e"). 

Thus  the  loss  of  kinetic  energy  is  (1—  c")  times  the  original  kinetic 
energy  relative  to  the  center  of  gravity.  If  the  bodies  are  perfectly 
elastic,  e  =  1,  so  that  there  is  no  loss  of  energy ;  while  if  (3  =  0,  the 
original  energy  relative  to  the  center  of  gravity  is  all  lost. 

Impact  of  Two  Smooth  Spheres 

202.  Let  us  apply  the  principles  just  explained  to  determining 
the  motion,  after  impact,  of  two  smooth  spheres. 

At  the  moment  of  impact  let  A,  B  be  the  centers  of  the  two 
spheres,  so  that  the  line  AB  is  the  common  normal  to  the  surfaces 
at  the  point  of  impact  C. 


IMPACT  OF  TWO  MOVING  BODIES  247 

As  before,  let  the  velocities  along  AB  before  impact  be  u,  u\ 
these  both  being  measured  in  the  direction  AB,  and  let  the 
velocities  in  the  same  direction  after  impact  be  v,  v'.  Then  we 
have,  by  the  conservation  of  momentum  along  AB, 

mu  +  m'u'  —  mv  +  7n'v' , 
and,  by  Newton's  law,       v  —  v'  =  —  e  (u  —  u'). 

From  these  equations  (88)  and  (89)  follow  as  before. 


Fig.  128 

If  the  velocities  of  A  before  impact  make  angles  a,  a'  with  AB 

as  marked  in  the  figure,  the  tangential  velocities  of  A  before  and 

after  impact  are  ,  ,        , 

^  u  tan  a,  —  v  tan  a, 

so   that,   since   the    tangential    velocities    remain   unaltered,   we 

must  have  ,        ,  , 

V  tan  a'  =  —  u  tan  a ; 

while  similarly,  from  the  motion  of  B, 

r'tan/S'=  — w'tanyS. 

Thus  equations  (88)  and  (89)  become 

^    ,  mw  +  m'u'  —  em'(u  —  u') 

cot  a'  = --^ '-  cot  a, 

{m  -\-m)u 

cot  ^'= — ■ — —^ '-  cot  A 

(m  +  m)u' 

giving  a',  /S'  in  terms  of  the  initial  motion. 


248  MOTION  OF   SYSTEMS  OF  PAETICLES 

If,  as  iu  the  game  of  billiards,  the  spheres  are  of  equal  mass  and 
the  second  sphere  is  originally  at  rest,  we  take  m  =  m',  u'  =  0,  and 

obtain  cota' =  -  ^{1- e)Gota,         /3' =  0. 

Thus  B  starts  mto  motion  along  the  line  of  centers,  as  it  obviously 
must  since  the  forces  which  set  it  in  motion  act  along  this  line. 

Since  e  is  always  less  than  unity  and  a  is  necessarily  acute,  cot  a' 
must  be  negative,  so  that  a'  will  be  obtuse.  If  e  =  1,  then  a'  =  90°. 
Thus  if  the  spheres  were  perfectly  smooth  and  perfectly  elastic,  A 
would  move  at  right  angles  to  the  line  of  centers  after  impact ;  its 
motion  would  be  the  same  as  if  it  had  impinged  on  a  perfectly 
smooth  and  inelastic  plane. 

ILLUSTRATIVE   EXAMPLE 

A  row  of  similar  coins  is  placed  on  a  rouyh  table,  the  coins  being  at  equal  dis- 
tances apart  and  in  a  straight  line.  The  first  coin  is  projected  along  this  line  so  as 
to  impinge  directly  on  the  second.    Find  the  resulting  motion. 

Let  e  be  the  coefficient  of  elasticity  for  an  impact  between  the  two  coins,  and 
ft  the  coefficient  of  friction  between  the  coins  and  the  table.  Let  m  be  the  mass 
of  each  coin,  and  d  the  distance  between  the  nearest  points  of  two  adjacent  coins. 

The  normal  reaction  between  a  coin  and  the  table  is  mg,  so  that  the  f  rictional 
force  opposing  the  coin's  motion  is  fimg,  and  the  retardation  produced  is  /xg. 
Thus  if  a  coin  is  started  from  its  original  position  with  a  velocity  V,  its  velocity 
on  reaching  the  next  coin  is  reduced  Xo  u,  where 

F2  -  «2  =  2  fj.gd.  (a) 

We  now  have  two  coins  of  equal  mass  impinging  with  velocities  u,  0.  Their 
velocities  after  impact,  say  v,  v',  are  given  by  the  equations 

V  —  v'=  —  eu  (Newton's  law), 

V  +  v'=  u  (conservation  of  momentum). 

Thus  v=lu(l~e), 

v'=  i  w  (1  +  e). 

After  impact  the  coin  originally  in  motion  has  a  velocity  v,  and  is  retarded 
by  a  f  rictional  retardation  ng.  It  accordingly  comes  to  rest,  if  it  does  not  collide 
again  in  the  meantime,  after  a  distance  s  given  by 

v^=  2  ugs, 

s  =  -'"^='''^'-'^\  \b) 

2g  Sg 

while  the  coin  which  has  been  started  into  motion  sets  off  with  a  velocity 

v'-  lu{l  +  e).  (c) 


ILLUSTRATIVE  EXAMPLE  '  249 

The  coin  before  this  started  into  motion  with  a  velocity  V  given  by  equa- 
tion (a).  Eliminating  u  from  equations  (a)  and  (c),  we  find  as  the  relation 
between  successive  velocities  of  starting 

F2  =  2/x^d  +  -if^,,  {d) 

(1  +  ey- 

a  difference  equation  with  constant  coefficients. 

If  e  =  1,  we  notice  from  equation  (6)  that  s  =  0,  so  that  each  coin  remains 
absolutely  at  rest  after  striking  the  coin  next  in  front  of  it  —  it  transmits  the 
whole  of  its  momentum  to  this  coin.    Also  from  equation  (a), 

F2-  v2-  2,jLgd. 

After  the  momentum  has  been  transmitted  over  the  space  between  n  coins, 
the  value  of  the  square  of  the  velocity  is  reduced  by  2  nfxgd.  Thus  at  any  point 
the  velocity  of  a  moving  coin  is  that  which  would  be  possessed  by  a  coin  which 
had  been  started  with  velocity  F,  and  made  to  move  over  a  distance  equal  to  all 
the  intervals  between  the  coins  over  which  the  motion  has  been  transmitted. 

If  d  =  0,  so  that  the  coins  were  originally  in  contact,  we  have 

F  =  ^. 

1  +  e 

Thus,  if  there  are  n  coins,  the  )ith  coin  will  start  in  motion  with  a  velocity 

\  2  ; 

EXAMPLES 

1.  Hailstones  are  observed  to  strike  the  surface  of  a  frozen  lake  in  a  direction 
making  an  angle  of  30  degrees  with  the  vertical,  and  to  rebound  at  an  angle  of 
60  degrees.    Assuming  the  contact  to  be  smooth,  find  the  coeflBcient  of  elasticity. 

2.  If  the  hailstones  of  the  last  question  rise  after  impact  to  a  height  of  2  feet, 
find  the  velocity  with  which  they  originally  struck  the  ground. 

3.  In  the  last  question  find  the  height  to  which  the  hailstones  will  rise  in 
their  second  rebound  from  the  ice. 

4.  A  ball  is  dropped  on  to  a  horizontal  floor  and,  after  rebounding  twice, 
reaches  a  height  equal  to  half  that  from  which  it  was  dropped.  Find  the 
coeflficient  of  elasticity. 

5.  A  bullet  strikes  a  rough  target  at  45  degrees,  and  rebounds  at  the  same 


angle.    Show  that 


1-e 


6.  A  shot  firecl  from  a  distance  a  strikes  a  target  at  right  angles,  and 
rebounds.  Show  that  it  will  fall  at  a  distance  ae  from  the  target  (neglecting 
the  resistance  of  the  air). 

7.  A  sphere  of  mass  m  collides  with  a  second  sphere  of  mass  m',  the  contact 
between  them  being  smooth,  and  their  paths  after  collision  are  observed  to  be 
at  right  angles.    Prove  that  m  =  em\ 


250  ^lOTIOX  OF   SYSTEMS   OF  PARTICLES 

8.  Two  billiard  balls  stand  at  rest,  and  a  third  ball  is  made  to  strike 
them  simultaneously,  and  is  observed  to  remain  at  rest  after  the  impact. 
Show  that  e  =  |. 

9.  A  particle  is  projected  from  a  point  on  a  smooth  horizontal  plane,  with 

velocity  V  at  an  elevation  a,  and  after  striking  the  plane  rebounds  time  after 

2  V  sin  fT 
time.    Show  that  its  total  time  of  flight  is  ,  and  that  its  total  range 

.    F-  sin  2  cr  y\/ 

IS 

9{l-e) 

10.  A  player  stands  at  a  horizontal  distance  d  from  a  wall,  and  throws  a 

ball  towards  the  wall  at  an  inclination  a  to  the  horizontal.    Show  that  if  it  is 

to  return  to  him  after  bouncing,'  he  must  throw  it  with  a  velocity  V  given  by 

y.2-  {^  +  e)9d 


2  e  cos  a  (sin  a  —  /x  cos  a) 

where  e,  /x  are  coefficients  of  elasticity  and  friction. 

11.  In   the    last   question   consider  the  cases  of  (a)  e  =  0,   (6)  /i  =  tana, 
(c)  fx  >  tan  a. 

GENERAL   EXAMPLES 

1.  A  2>article  is  placed  on  the  face  of  a  smooth  wedge  which  can  slide 
on  a  horizontal  table  ;  find  how  the  wedge  must  be  moved  in  order  that 
the  particle  may  neither  ascend  nor  descend.  Also  find  the  pressure  between, 
the  particle  and  the  wedge. 

2.  It  is  required  to  run  trains  of  100  tons  on  a  level  electric  railway, 
with  stations  half  a  mile  apart,  at  an  average  speed  of  12  miles  an  hour, 
including  half  a  minute  stop  at  each  station.  Prove  that  the  electric  loco- 
motives must  weigh  at  least  an  additional  8  tons,  taking  a  coefficient  of 
friction  of  I,  and  supposing  the  trains  fitted  with  continuous  brakes. 
(Neglect  passive  resistances.) 

Prove  that  the  railway  can  be  worked  by  gravity,  if  the  line  is  curved 
downward  between  the  stations  to  a  radius  of  about  46,000  feet ;  and 
that  the  dip  between  the  stations  will  be  about  20  feet,  the  inclines 
at  the  stations  about  1  in  33,  and  the  maximum  velocity  about  23|-  miles 
an  hour. 

3.  A  cylinder  of  height  h  and  diameter  d  stands  on  the  floor  of  a  rail- 
way car,  which  suddenly  begins  to  move  with  acceleration  /.  Show  that 
the  cylinder  will  only  remain  at  rest  relative  to  the  car  if /is  less  than 
both  fxg  and  (l(//h. 

4.  If  a  circular  hoop  is  projected,  spinning  steadily  without  wobbling, 
prove  that  the  center  describes  a  parabola,  and  that  the  tension  of  the  rim 
is  the  weight  of  a  length  v-/g  of  the  rim,  where  v  denotes  the  rim  velocity 
relative  to  the  center  of  the  hoop. 


EXAMPLES  251 

5.  A  uniform  chain  G  feet  long,  liaving  a  mass  of  2  pounds  per  foot,  is 
laid  in  a  straight  line  along  a  rough  horizontal  table,  for  which  the  coeffi- 
cient of  friction  is  h,  a  portion  hanging  over  the  edge  of  the  table  so  that 
slipping  is  just  about  to  occur.  If  a  slight  disturbance  sets  the  chain  slip- 
ping, find  the  tension  at  the  edge  of  the  table  when  z  feet  have  slipped  off. 

6.  Two  equal  balls  A,  B,  each  of  mass  m,  are  at  a  distance  a  apart. 
An  impulse  /  acts  on  .4  in  the  direction  AB,  and  a  constant  force  F  acts 
on  B  in  the  same  direction.    Show  that  A  will  not  overtake  B  if 

72  <  2  aFm. 

7.  A  bullet  weighing  one  ounce  is  fired  with  a  velocity  of  1200  feet  per 
second  at  an  elevation  of  1  degree  so  as  to  hit  a  bird  weighing  2\  pounds 
when  the  bullet  is  at  the  highest  point  of  its  path.  Supposing  the  bird  to 
have  been  at  rest  when  hit  and  afterwards  to  fall  with  the  bullet  embedded 
in  it,  find  how  far  from  the  point  of  firing  the  bird  will  fall. 

8.  If  a  bullet  weighing  iv  pounds  is  fired  with  velocity  v  at  a  body 
weighing  W  pounds,  advancing  with  velocity  V,  prove  that  the  body  will 
retain  the  velocity 

WV  —  wo  -.         MJ     .  , 

J  or    V (v  ■- u), 

W+w  W^  ^ 

according  as  the  bullet  is  embedded,  or  perforates  and  retains  a  velocity  u. 
Calculate  the  energy  liberated,  and  thence  infer  the  average  resistance  of 
the  body  from  the  length  perforated  by  the  bullet. 

9.  A  pile  is  being  driven  in  by  repeated  impacts  of  a  falling  weight. 
How  does  the  extent  to  which  the  pile  is  driven  in  by  each  blow  depend 
(a)  on  the  magnitude  of  the  weight,  and  (b)  on  the  height  to  which  it  is 
raised  before  being  released  ? 

If  the  weight  be  1  ton,  and  the  height  from  which  it  falls  be  10  feet, 
and  the  pile  be  driven  in  a  tenth  of  an  inch,  find  the  resistance  in  tons. 

10.   An  inelastic  pile,  of  mass  vi  pounds,  is  driven  vertically  into  the 

ground  a  distance  of  a  feet  at  each  blow  of  a  hammer  of  mass  M  pounds, 

which  falls  vertically  through  h  feet.    Show  that  the  weight  which  would 

have  to  be  placed  on  the  top  of  the  pile  to  drive  it  slowly  into  the  ground 

would  be  ,.,, 

Jl  -I pounds. 

(J/  +  )n)a 

11.  A  hammer  head  of  W  pounds,  moving  with  a  velocity  of  v  feet  a 
second,  strikes  an  inelastic  nail  of  lo  pounds  fixed  in  a  block  of  j\I  pounds 
which  is  free  to  move.  Prove  that  if  the  mean  resistance  of  the  block  to 
penetration  by  the  nail  is  a  force  ot.R  pounds,  then  the  nail  will  penetrate 
each  blow  a  distance,  in  feet, 

MW^ if_ 

(M  +  W+  w)  (W+  w)  2  (jR 


252  MOTION  OF   SYSTEMS   OF  PARTICLES 

12.  In  the  system  of  pulleys  described  in  §  128,  show  that  if  P  is  a  weight 
which  is  not  equ&\  to  W/n,  the  acceleration  produced  in  the  weight  W  will  be 

nP  -  W 

13.  Two  masses  m,  m'  connected  by  an  elastic  string  are  placed  on  a 
smooth  horizontal  table,  the  masses  being  at  rest  and  the  string  unstretched. 
A  blow  of  impulse  P  is  given  to  the  first  mass,  in  the  direction  away  from 
the  second.  Show  that  when  the  string  is  again  unstretched,  the  velocity 
of  the  second  mass  is  ^  „ 


m  +  m' 


14.  Three  equal  particles  are  tied  at  the  ends  and  middle  point  of  an 
inextensible  string,  which  is  placed,  fully  extended,  on  a  smooth  table. 
The  middle  particle  is  jerked  into  motion  in  the  direction  towards  and 
perpendicular  to  the  line  joining  the  other  two.  Find  the  loss  of  energy 
when  the  other  particles  are  jerked  into  motion. 

15.  A  coal  train  consists  of  a  number  of  similar  trucks  hauled  by  an 
engine  whose  weight  is  just  equal  to  that  of  three  trucks.  The  train  is  at 
rest  on  a  level  track,  the  couplings,  which  are  of  equal  length,  being  all 
equally  slack.  The  engine  then  begins  to  move  with  a  constant  tractive 
force,  and  each  truck  is  jerked  into  motion  as  its  coupling  tightens.  Show 
that  the  speed  of  the  engine  will  be  greatest  just  before  the  tenth  jerk 
occurs. 

16.  Snow  is  evenly  spread  over  a  roof.  If  a  mass  commences  to  slide, 
clearing  away  a  path  of  uniform  breadth  as  it  goes,  prove  that  its  acceler- 
ation is  constant,  and  equal  to  a  third  of  that  of  a  mass  sliding  freely 
down  the  roof. 

17.  A  heavy,  perfectly  flexible  uniform  string  hanging  vertically  with 
its  lowest  i^oint  at  a  height  h  above  an  inelastic  horizontal  plane  is  sud- 
denly allowed  to  fall  on  to  the  plane.  Show  that  the  pressure  on  the  table 
when  a  length  x  of  the  string  has  fallen  on  to  the  table  is 

(3x4-  2h)mfj. 

18.  Show   that    if    two    equal    balls   impinge   directly   with  velocities 

^  V  and  —  F,  the  former  will  be  reduced  to  rest. 

1  —  e 

19.  Show  that  the  mass  m  of  a  sphere  which  must  be  interposed  between 
a  sphere  of  mass  M  at  rest  and  one  of  mass  M'  moving  directly  on  to  it 
with  velocity  V,  in  order  that  the  former  may  acquire  the  greatest  possible 
velocity  from  the  impact,  will  be  VJlW' ,  and  that  the  velocity  acquired 
will  be  M'V{\+ey 

M  +  M'  +  2  m 


EXAMPLES  253 

20.  Prove  that  an  elastic  ball,  let  fall  vertically  from  a  height  of  h  feet 
on  a  hard  pavement,  and  rebounding  each  time  vertically  with  e  times  the 
striking  velocity,  will  have  described 

1  +  e2     ,     ,    .     1  +  e     /2A  , 

Ii  leet,  m -kI —  seconds 

l-e2  \-e\  g 

before  the  reboiinds  cease. 

Work  this  out  for  A  =  1 ,  e  =  |. 

21.  A  ball  is  dropped  from  the  top  of  a  tower,  height  7*,  and  at  the  same 
time  another  ball  of  equal  weight  is  j^rojected  upwards  with  the  velocity 
V2  gh  from  the  base  of  the  tower  and  collides  directly  with  the  falling  ball. 
If  the  coefficient  of  restitution  be  e,  prove  that  the  falling  ball  will,  in 

the  rebound,  rise  to  a  height  short  of  the  top  of  the  tower  by  -  (1  —  e'-). 

4 

22.  A  boy  standing  on  a  railway  bridge  lets  a  ball  fall  on  the  horizontal 
roof  of  a  car  passing  vmder  the  bridge  at  15  miles  an  hour.  If  ^u  =  ^,  e  =  f 
between  the  roof  and  the  ball,  find  the  least  height  of  the  boy's  hand  above 
the  roof  in  order  that  the  second  rebound  of  the  ball  may  be  from  the 
same  point  of  the  roof  as  the  first. 

If  the  boy's  hand  is  at  a  gi-eater  height  than  this,  what  will  happen? 

23.  A  perfectly  elastic  particle  is  projected  so  as  to  strike  the  inside  of 
a  surface  of  revolution  of  which  the  axis  is  a  given  vertical  line.  Show 
that  the  vertices  of  all  the  parabolas  described  after  successive  rebounds 
lie  on  a  surface  of  which  the  shape  is  independent  of  that  of  the  surface 
of  revolution. 

24.  Prove  that  in  order  to  jaroduce  the  greatest  possible  deviation  in  the 
direction  of  motion  of  a  smooth  billiard  ball  of  diameter  a  by  impact  on 
another  equal  ball  at  rest,  the  former  must  be  projected  in  a  direction 
making  an  angle 


with  the  line,  of  length  c,  joining  the  two  centers. 

25.  A  pendulum  hangs  with  its  bob  just  in  contact  with  a  smooth  verti- 
cal plane.  The  bob  is  drawn  aside  until  it  is  .5  inches  higher  than  it  was, 
and  is  then  released  so  as  to  strike  the  plane  normally  ;  and  on  the  first 
rebound  it  rises  vertically  through  4  inches.  What  would  have  been  the 
vertical  rise  on  rebound  if  the  pendulum  had  been  drawn  aside  through 
the  same  angle,  but  so  that  the  bob  strikes  at  an  angle  of  60  degrees  with 
the  normal? 


CHAPTEE   X 
MOTION  OF  A  PARTICLE  UNDER  A  VARIABLE  FORCE 

203.  In  almost  all  the  cases  of  motion  of  a  particle  which 
have  so  far  been  considered,  the  forces  acting  on  the  particle 
have  remained  constant  throughout  the  whole  of  the  path,  so 
that  the  acceleration  of  the  particle  has  been  constant.  We  pro- 
ceed now  to  consider  the  motion  of  a  particle  which  is  acted 
upon  by  forces  wliich  vary  from  point  to  pomt  of  the  path  of 
the  particle. 

These  problems  fall  into  two  classes,  according  to  whether  the 
jDath  described  by  the  particle  is  or  is  not  given  as  one  of  the  data 
of  the  problem.  The  former  class  is  the  simplest  and  is  considered 
first.  It  includes  such  cases  as  the  motion  of  a  pendulum,  in  which 
the  "  bob  "  of  the  pendulum  is  constrained  to  describe  a  circle  by 
the  mechanism  of  suspension  of  the  pendulum,  as  also  that  of  the 
motion  of  a  bead  on  a  wire,  in  which  the  bead  is  compelled  to 
describe  the  path  marked  out  for  it  by  the  wire. 

Equations  of  Motion 

204.  Let  s  denote  the  distance  described  by  the  particle  along 
its  path  at  any  instant  t,  tliis  distance  being  measured  from  any 

„    fixed  point  0  on  the  path.    The  velocity  along 

ds 
the  path  is  then  —  •    Calling  this  v,  the  accelera- 


FiG.  129 


.     dv        d'^s 
tion  IS  -—  or  — -T- 

dt       de 


We  can  also  obtain  a  value  for  the  accelera- 
tion fro^  a  knowledge  of  the  forces  acting.  To  find  the  acceler- 
ation, we  must  resolve  all  the  forces  which  act  on  the  particle  in 
the  direction  of  the  path.     If  *S'  is  the  component  of  force  in  this 

254 


EQUATIONS  OF  MOTION  255 

direction,  the  equation  of  motion  of  the  particle,  by  the  second 
law  of  motion,  will  be 

S  =  ^±,  (90, 

or  S=m-^-  •  (91) 

df  ^     ' 

We  shall  suppose  the  field  of  force  to  be  permanent,  so  that  the 
quantity  *S'  may  be  supposed  to  depend  only  on  the  position  occu- 
pied by  the  particle  on  its  path,  and  not  on  the  instant  at  which 
it  arrives  there.  In  other  words,  >S  is  a  fimction  of  s  but  not  of  t. 
Equation  (91)  is  a  differential  equation  connecting  s  and  t\  if  we 
can  solve  this  equation,  we  shall  have  a  full  knowledge  of  the 
motion  of  the  particle  provided  its  path  is  known. 

The  equation  is  a  differential  equation  of  the  second  order,  but 
can  easily  be  transformed  into  one  of  the  first  order.    For 


d's      dv      dv  ds 
df  ~dt~dsdt~ 

dv 
ds 

so  that  the  equation  can  be  written 

S  =  mv  —-■ 
ds 

Smce  *S  is  a  function  of  s,  tliis  equation  can  be  integrated  with 
respect  to  s,  so  that  we  obtain 

C+  fsds  =  lmv\  (92) 

where  C  is  a  constant  of  integration. 

Since  v  is  equal  to  -^  >  this  equation  can  be  written  in  the  form 


ds 
dt 


^i{''+j'^)'  <«'> 


which  is  an  equation  of  the  first  degree.    If  this  can  be  solved,  the 
solution  of  the  problem  is  complete. 


256  MOTION  UNDER  A  VAKIABLE  FORCE 

We  notice  that  the  right  hand  of  equation  (92)  is  the  kinetic 
energy  of  the  particle.    Also,  since  the  force  opposing  the  motion 

of  the  particle  along  its  path  is  —  S,  its  potential  energy  is  —  j  S  ds. 

Thus  equation  (92)  expresses  that  the  sum  of  the  kinetic  and 
potential  energies  remains  constant  —  it  is  the  equation  of  energy 
for  the  motion  of  the  particle.  From  a  knowledge  of  the  total 
energy  at  any  instant  of  the  particle's  'motion,  we  can  determine 
the  constant  C,  and  can  then  proceed  to  integrate  equation  (93), 
if  possible. 

ILLUSTRATIVE   EXAMPLE 

Assuming  that  the  value  of  gravity  falls  off  inversely  as  the  square  of  the  dis- 
tance from  the  earth'' s  center^  determine  the  motion  of  a  projectile  fired  vertically 
into  the  air,  the  diminution  of  gravity  being  taken  into  account. 

Let  a  be  the  radius  of  the  earth,  and  g  the  value  of  gravity  at  the  surface. 

Then,  at  a  distance  r  from  the  earth's  center,  tlie  value  of  gravity  will  be  - — 

Since  the  particle  moves  along  a  radius  drawn  through  the  center  of  the 

earth,  we  may  measure  all  distances  from  the  earth's  center,  and  the  distance 

r  from  the  earth's  center  may  replace  the  coordinate  s  of  §  204.    The  value  of 

7ncfa^ 
the  force  S  resolved  along  the  path  is ~ — ,  so  that  the  equation  of  motion  is 

mga^  _      d'^r 
7^~'    d^' 

The  equation  of  energy,  as  in  equation  (92),  is 

72 


0-1    '—  dr  =  I  mv^, 
J     r^  ' 


or  C  +  ^^^^"=:imv2.  (a) 

r 

Let  us  suppose  that  the  particle  was  projected  from  the  earth's  surface  with 
velocity  V.  Putting  r  —  a  in  equation  (a),  the  value  of  v  must  be  F,  so  that 
we  must  have 

C  +  mga  =  h  mV^,  (6) 

and  this  equation  determines  the  value  of  C.    Eliminating  C  from  equations  (a) 
and  (6),  we  obtain 

ga(^l-^^  =  l(V^-v% 


ILLUSTRATIVE  EXAMPLE  257 

giving  the  velocity  at  any  point  in  tlie  form 


v^^V^-2ga(l-fj. 


Since  v  =  — ,  this  equation  becomes 
dt 


=  yjv^-2ga(l-fj, 


(c) 


id) 


/civ 
— ■•  (e) 


^V^-2ga  +  ^-^ 


On  performing  the  integration,  we  can  find  the  time  required  to  describe  any 
portion  of  the  path.    Let  us  first  consider  the  different  types  of  solution. 
We  see  from  equation  (c)  that  v  vanishes  when 

2ga'^ 


2ga-  F2 


so  that  if  V^  <  2  ga,  there  is  a  positive  value  of  r,  intermediate  between  +  a 
and  +00,  for  which  the  velocity  vanishes.    Thus  if  V^<2ga,  the  projectile 

goes  to  the  point  at  distance  — — ,  and  then  falls  back  on  to  the  earth.    If 

2ga  —  V- 

V^  >  '2ga,  we  find  that  there  is  no  positive  value  of  r  for  which  v  vanishes,  so 

that  the  particle  goes  to  infinity  :  it  escapes  from  the  earth  altogether. 

When  V^  =  2ga,  the  velocity  vanishes  at  infinity;  thus  the  particle  just 
escapes  from  the  earth's  attraction,  but  is  left  with  zero  velocity.  Its  kinetic 
energy  of  projection  is  just  used  up  in  overcoming  the  earth's  attraction. 

Let  us  consider  first  the  special  case  in  which  V'^  —  2  ga.  We  find  that  equa- 
tion (e)  reduces  to 


t 


/dr 


dr 


+  C", 


where  C  is  a  new  constant  of  integration. 

If  we  measure  time  from  the  instant  of  projection,  we  must  have  t  =  0  when 
r  =  a,  so  that 


and  on  eliminating  C, 


0  =  \^  +  C% 
3\2g 


t  = -^{r^-ai).  (f) 

3  a  V2gr 


258  MOTION   UNDER  A  VARIABLE  FORCE 

In  the  case  in  which  V^  >  2  ga,  we  obtain  after  integration  of  equation  (e). 


t  =  - 


ga- 


2ga 


V- 


2ga  + 


2  ga'^ 


■Vv^-2ga 


log— p= 


-W-  -2ga 


2ga^ 


2ga  +  -^-~  +  VF2  -2ga 


26 


v^ 


2ga  + 


2ga~ 


y  +  c', 


(?) 


J 


where  C  is  a  new  constant  of  integration.    If  the  time  is  to  be  measured  from 
the  instant  at  which  the  particle  is  projected,  we  must  have  ^  =  0  when  r  =  a^ 

so  that  

1  ,      V-^/V^-2ga      2br  , 

los -^    +  —r-  y  +  C", 


0  = 


V^-2ga  I  VF2-2 


ga 


F  +  VF2-2 


ga 


and  on  eliminating  C,  we  can  again  obtain  the  time  required  to  describe  any 
portion  of  the  path.  The  case  in  which  V^  <2ga  can  be  treated  in  a  similar 
manner.    This  is  left  as  an  example  for  the  student. 


EXAMPLES 

1.  In  the  foregoing  illustrative  example,  suppose  that  V'^<2ga,  and  find 
(a)  the  greatest  height  reached  ; 

(6)  the  time  of  flight  of  the  particle. 

2.  A  meteorite  falls  on  to  the  earth.  Assuming  it  to  start  from  infinity  with 
zero  velocity,  and  to  fall  directly  on  to  the  earth,  find  the  velocity  with  which 
it  reaches  the  earth's  surface,  and  the  time  taken  to  fall  to  the  earth's  surface 
from  a  point  distant  r  from  the  earth's  center. 

3.  A  particle  falls  from  distance  a  into  a  center  of  force  which  attracts 
according  to  the  law  fi/r^.  Show  that  the  average  velocity  on  the  first  half  of 
the  path  Ls  to  the  average  velocity  on  the  second  half  in  the  ratio 

TT  -  2  :  TT  +  2. 

4.  Find  the  time  of  falling  to  a  center  of  force  which  attracts  according 
to  the  law  /j-r"  K 

5.  A  particle  moves  in  a  straight  line  from  a  distance  a  to  a  center  of  attrac- 
tion towards  which  the  force  is  —   Show  that  the  time  required  to  reach  the 

center  is  ^ 

a- 

6.  A  particle  begins  to  move  from  a  distance  a  towards^fixed  center  which 
repels  according  to  the  law  fxr.  If  its  initial  velocity  is  V/xa,  show  that  it  will 
continually  approach  the  fixed  center,  but  will  never  reach  it. 


THE   SIMPLE  PENDULUM 


259 


The  Simple  Pendulum 

205.  One  of  the  most  important  cases  of  a  variable  force  arises 
in  the  motion  of  a  simple  pendulum.  To  obtain  a  first  approxima- 
tion, we  can  suppose  that  the  whole  weight  of  the  pendulum  is 
concentrated  in  the  bob,  which  may  be 
treated  as  a  particle,  and  that  this  is  sus- 
pended from  a  fixed  point  by  a  weight- 
less string  or  rod  so  that  it  is  constrained 
to  move  in  a  vertical  circle. 

Let  s  denote  the  distance  along  this 
circle  described  by  the  particle,  this  dis- 
tance being  measured  from  the  lowest 
point  0.  Let  the  angle  PCO  between 
the  string  and  the  vertical  be  denoted 
by  0,  so  that  s  =  ad.  The  forces  actmg 
on  the  particle  consist  of  its  weight  and  the  tension  of  the 
string.  The  latter  has  no  component  in  the  direction  in  which 
the  particle  moves.  The  former  has  a  component  —  mg  sm  6. 
Thus  the  equation  of  motion  is 


cPs 
di 


-  =  -g  sin  6, 


(94) 


where  6  =  s/a. 

206.  This  equation  cannot  be  solved  by  elementary  mathe- 
matics, except  m  the  simple  case  in  which  the  angle  6  is  small, 
—  i.e.  the  case  in  which  the  pendulum  never  swings  through 
more  than  a  small  angle  from  the  vertical.  Confining  our  atten- 
tion to  this  case,  we  may  replace  sin  6  by  6,  and  6  by  s/a,  and 
write  the  equation  of  motion  in  the  form 


Tlius  the  acceleration  of  the  bob  of  the  pendulum  is  proportional 
to  its  distance  from  0,  and  is  towards  0. 


260  MOTION   UNDER  A  VARIABLE  FORCE 

Writing  the  equation  in  the  form 
dv  (g 


ds  \a, 


and  integrating  with  respect  to  s,  we  obtain 

ri,'^=C-[^s\  (95) 


Clearly  the  constant  C  must  be  positive,  and  the  velocity  will 
vanish  as  soon  as  s  reaches  a  value  such  that 


Let  us  denote  the  two  values  of  s  which  satisfy  this  equation  by 
±  s„ ,  then  the  motion  of  the  bob  is  clearly  confined  witliin  two 
points  at  distances  s^  from  the  point  0  on  opposite  sides.  Thus 
we  may  call  s^  the  amplitude  of  the  swing. 

Eeplacing  C  by  (-)■Su^  equation  (95)  becomes 

^'=|(-^o'-A  (96) 


ds       \g ,  o      OS 

so  that  —  =  -yl-  (s;  —  s'), 

at         ya 

and  the  integral  of  tliis  equation  is 

t  = 


'/°"ur'' 


where  e  is  a  constant  of  mtegration. 
This  equation  gives 


}y^^^^-^' 


cos 


so  that  s  =  Sq  cos  ]  -y  -  {t  —  e) 


THE  SIMPLE  PENDULUM  2G1 

This  equation  contains  the  complete  solution  of  the  problem. 
We  notice  that  the  values  of  s  continually  repeat  at  intervals  of 
time  L  for  which 


V 


^-4  =  2.. 


Thus  the  motion  of  the  pendulum  repeats  itself  indefinitely. 
The  interval  between  two  instants  at  which  the  pendulum  is  in 
the  same  position,  namely  t^,  given  by 

is  called  the  period. 

207.  Seconds  pendulum.  To  construct  a  pendulum  which  is  to 
beat  seconds,  we  choose  a  so  that  ^^  shall  be  equal  to  two  seconds, 
for  a  seconds  pendulum  is  one  which  takes  one  second  to  move 
from  left  to  right,  and  then  one  second  more  to  move  from  right 
to  left.    Thus  we  must  have 

^  =  1. 

In  foot-second  units  we  may  take  g=  32.19  for  London,  and 

so  obtain 

a  =  39.14  inches, 

as  the  length  of  the  seconds  pendulum  at  London. 

We  notice  that  the  period  of  a  pendulum  varies  as  the  square 
root  of  its  length.  Thus,  for  a  pendulum  to  beat  half-seconds,  its 
length  would  have  to  be  only  a  quarter  of  that  of  the  seconds 
pendulum,  and  therefore  9.78  inches  at  London. 

Since  g  varies  from  point  to  point  on  the  earth's  surface,  the 
length  of  the  seconds  pendulum  will  also  vary.  If  we  observe 
the  length  of  a  pendulum  and  also  measure  its  period  with  a 
chronometer,  we  shall  be  able  to  calculate  the  value  of  g  at 
the  place  at  which  the  experiment  is  performed ;  in  fact,  this 
method  affords  the  easiest  and  most  accurate  way  of  obtaining 
the  value  of  g  at  any  point  of  the  earth's  surface. 


2G2  MOTION    UNDEK  A  VARIABLE  FORCE 


ILLUSTRATIVE  EXAMPLE 

A  pendulum  which  beats  seconds  accurately  at  New  York  is  found  to  gain,  2 
seconds  a  day  when  taken  to  Philadelj)hia.  Compare  the  values  of  g  at  Philadel- 
phia and  New  York. 

At  Philadelphia  the  pendulum  makes  24  x  (00)2  _|.  2  beats  in  24  x  (G0)2 
seconds.    Thus  the  time  of  a  beat  is 

24  X  (60)2 

seconds. 


24  X  (60)2  +  2 

and  this  is  equal  to  7r\/-,  where  a  is  the  length  of  the  pendulum  and  g  is  the 
value  of  gravity  at  Philadelphia.  If  go  denote  the  value  of  gravity  at  New  York, 
go  =  TT-a, 


we  have 


,    r24  X  (60)2  +  2"| 

g  =  TT-a    ^^ 

L    24  X  (60)2     J 


=  7r2a  I  1  + 


—  I  approximately, 
)-}  •     ' 


24  X  (60)'- 

so  that  g  =  go  -^  (^  -\ 1 

'^      •"        \        24  X  (60)2/ 

=  go(l-  2jioo)  approximately. 

Thus  gravity  is  less  at  Philadelphia  than  at  New  York  by  about  one  part  in 
21,600. 

EXAMPLES 

1.  Calculate  the  length  of  a  pendulum  to  beat  time  to  a  march  of  100  paces 
a  minute. 

2.  A  pendulum  which  beats  seconds  in  London  requires  to  be  shortened  by 
one  thousandth  of  its  length  if  it  is  to  keep  time  in  New  York.  Compare  the 
values  of  gravity  at  London  and  New  York. 

3.  The  value  of  ^  at  a  point  on  the  earth's  surface  in  latitude  \  is 

g  =  go{l  -  .00257  cos  2  X), 

wliere  go  —  32.17  is  the  value  of  g  in  latitude  45  degrees.  Show  that  the  lati- 
tude in  which  a  short  journey  of  given  length  will  produce  the  greatest  error 
in  the  rate  of  a  pendulum  clock  is  latitude  45  degrees,  and  find  the  error  per 
mile  in  this  latitude.    (One  minute  of  latitude  =  6075  feet.) 

4.  In  a  building,  at  height  h  feet  above  the  ground,  the  value  of  gravity  is 

go  -  .0001  h, 

where  go  is  the  value  of  gravity  at  the  foot  of  the  building.  In  New  York, 
go  —  32.14.  Find  the  error  in  the  rate  of  a  pendulum  clock,  produced  by  taking 
it  from  the  ground  to  the  top  of  a  building  300  feet  high. 


SIMPLE  HARMONIC   MOTION  263 

5.  The  length  of  a  pendulum  which  makes  2  n  beats  per  day  is  changed 

nL 
from  Itol  +  L.    Show  that  the  pendulum  will  lose  —  beats  per  day  approxi- 
mately. 

6.  A  balloon  ascends  with  constant  acceleration,  and  reaches  a  height  of 
SGOO  feet  in  two  minutes.  Show  that  during  the  ascent  a  pendulum  clock  will 
have  lost  about  one  second. 

7.  A  ijendulum  of  length  I  is  adjusted  by  moving  a  small  part  only  of  the 
bob  of  the  pendulum,  this  being  of  mass  equal  to  one  nth  of  the  complete  bob. 
How  far  must  this  be  moved  to  correct  an  error  of  p  seconds  a  day  ? 


Simple  Harmonic  Motion 

208.  We  have  seen  that  throughout  the  motion  of  a  pendulum 
which  moves  so  that  its  maximum  inclination  to  the  vertical 
is  small,  the  acceleration  is  proportional  to  the  distance  from 
the  middle  pomt  of  its  path,  and  is  directed  towards  that  point. 
A  point  which  moves  m  this  way  is  said  to  move  with  simple 
harmonic  motion.  Thus  if  s  is  the  distance  from  a  fixed  pomt, 
of  a  point  which  moves  with  simple  harmonic  motion,  we  have 
an  equation  of  the  form 

dh 

de 
where  k  is  a  constant. 

Integrating,  we  obtain,  as  before  in  the  case  of  the  pendulum 
(cf.  equation  (96)), 

and  from  tliis  again 

s  =  Sq  cos  k  {t  —  e).  (97) 

The  constant  k  is  known  as  the  frequency  of  the  motion.  Thus 
the  frequency  of  a  simple  pendulum  is  \  —  • 

209.  A  simple  geometrical  interpretation  can  be  given  of  simple 
harmonic  motion,  and  this  enables  lis  to  obtain  a  complete  knowl- 
edge of  the  motion  without  any  use  of  the  integral  calculus  or  of 
the  theory  of  differential  equations.  In  fig.  131  let  the  arm  OP 
rotate  about  0  with  uniform  angular  velocity  k,  so  that  P  describes 
a  circle  of  radius  a  with  uniform  velocity  ka.    Let  a  perpendicular 


264 


MOTION   UNDER  A  VARIABLE   FORCE 


FN  be  drawn  from  P  to  a  fixed  diameter  AA'.    Then  we  shall  find 

that  the  pomt  iV  moves  backwards  and  forwards  on  the  line  AA' 

with  simple  harmonic  motion. 

The  acceleration  of  P  is,  by  §  12, 

an  acceleration  k^a  along  PO.    This 

can  be  regarded  as  compounded  of 

the  acceleration  of  P  relative  to  N, 

which  must  be  along  NP,  and  the 

acceleration    of   iV  relative    to    0, 

which    must    be    along  ON.    Thus 

the  acceleration  of  N  is  that  com- 

FiG.  131  ponent   of    the    acceleration   of  P 

which  is  in  the  direction  AA'.    This,  however,  is  known  to  be 

k^a  sin  d,   or   F  •  ON,   along  NO.     Putting   ON  =  s,   we   have   an 

acceleration  —  Jc^s  in  the  direction  in  which  s  is  measured,  namely 

ON.    Thus  the  point  N  moves  with  simple  harmonic  motion. 

This    geometrical  interpretation    of    simple   harmonic    motion 

enables  us  to  obtain  expressions  for  v  and  s  directly.    The  value  of 

s  is  ON,  or  a  cos  6.    Let  ^  =  e  be  an  instant  at  which  the  point  P 

was  passing  through  the  point  A'  in  its  motion  round  the  circle, 

then,  at  any  subsequent  instant  t,  the  time  since  P  was  at  A'  will 

he  t  —  €,  so  that  the  angle  described  by  OP  will  be  0  =  k{t  —  e). 

Thus  we  have 

s  =  ON=  a  cos  k  (t  -  e).  (98) 

This  is  the  same  result  as  is  contained  in  equation  (97).  We 
notice  that  the  amplitude  s^  of  the  motion  is  the  same  as  the 
radius  a  of  the  circle,  and  that  the  frequency  k  is  identical  with 
the  angular  velocity.  On  differentiating  equation  (98),  we  obtain 
at  once  for  the  velocity 


V  =  —  =  —  ka  sin  k  (t 
dt  ^^ 


^) 


This  result  can  also  be  obtained  by  resolving  the  velocity  ka  of 
the  moving  point  P  into  two  components,  along  and  perpendicular 


THE  CYCLOIDAL  PENDULUM  265 

to  AA'.    The  former  is  obviously  the  velocity  of  N  along  AA',  and 
it  is  at  once  seen  to  be  of  amount  —  Jm  sin  0,  or 

V  =  —  Jm  sin  k  {t  —  e) 
=  k  V«^  —  s',  as  before. 

In  this  motion,  as  in  the  motion  of  the  simple  pendulum,  the 

2  TT 

quantity  a  is  called  the  amplitude,  while  the  time  — —  after  which 
the  motion  repeats  itself  is  called  the  period. 

EXAMPLES 

1.  A  point  moves  ■with  simple  harmonic  motion  of  period  12  seconds,  and  has 
an  amplitude  of  5  feet.  Find  its  maximuoi  velocity,  and  find  its  position  and 
velocity  one  second  after  an  instant  at  which  its  velocity  is  a  maximum. 

2.  A  particle  moving  with  simple  harmonic  motion  of  period  t  is  observed 
to  have  a  velocity  v  when  at  a  distance  a  from  its  mean  position.  Find  its 
amplitude. 

3.  A  particle  is  free  to  move  along  a  line  AB  and  is  acted  on  by  an  atti'act- 
ive  force  directly  proportional  to  its  distance  from  a  point  P  in  AB,  and  con- 
sequently moves  with  simple  harmonic  motion.  Prove  that  its  average  kinetic 
energy  is  equal  to  its  average  potential  energy. 

4.  A  point  moving  with  simple  harmonic  motion  is  observ^ed  to  have 
velocities  of  3  and  4  feet  per  second  when  at  distances  of  4  and  3  feet  respec- 
tively from  its  mean  position.    Find  its  amplitude  and  period. 

5.  A  point  moves  with  simple  harmonic  motion  relative  to  one  frame,  and 
the  frame  itself  moves  with  simple  harmonic  motion  relative  to  a  second  frame, 
the  directions  of  the  two  motions  being  parallel,  and  their  periods  the  same. 
Show  that  the  motion  of  the  moving  point  relative  to  the  second  frame  is  simple 
harmonic  motion,  of  the  same  direction  and  period  as  that  of  the  frame. 

6.  A  weight  w  is  tied  to  an  elastic  string  of  natural  length  a  and  modulus  X, 
and  is  allowed  to  hang  vertically  in  equilibrium.  The  weight  is  now  pulled 
down  vertically  through  a  further  distance  h.  Show  that  on  being  set  free  it 
will  describe  simple  harmonic  motion  of  amplitude  a,  provided  this  does  not 
involve  the  string  ever  becoming  unstretched.    Find  the  period  of  the  motion. 

The  Cycloidal  Pendulum 

210.  AVe  have  seen  that  the  motion  of  a  simple  pendulum  is 
simple  harmonic  motion  only  so  long  as  the  amplitude  of  the 
motion  is  small.  It  is,  however,  possible  to  constrain  a  particle  to 
move  imder  gravity  m  such  a  way  that  its  motion  shall  be  simple 
harmonic  motion  no  matter  how  great  the  amplitude. 


266 


MOTION  UNDER  A  YARIA15LE  FORCE 


To  find  the  curve  in  wliicli  the  particle  must  be  constrained  to 
move,  let  us  go  back  to  equation  (94),  namely 


d's 


which  is  the  equation  of  motion  of  a  particle  constrained  to  move 
in  any  cur\'e,  provided  6  is  the  angle  which  the  tangent  to  the 
curve  at  a  distance  s  along  it  makes  with  the  horizontal.  For  this 
equation  to  represent  simple  harmonic  motion,  the  acceleration 
drs 


——  must  be  equal  to  —  Irs.    Thus  we  must  have 

(/  sm6  =  Ic^s, 


(99) 


so  that  sin  6  must  be  proportional  to  s. 

211.  Tliis  relation  expresses  a  property  of  the  cycloid, —  i.e.  of 
the  curve  described  in  space  by  a  point  on  the  rim  of  a  circle  which 
rolls  along  a  straight  line.  For,  in  fig.  132,  let  P  be  a  point  on 
a  cycloid  which  is  formed  by  a  circle  rolling  along  the  line  EF. 
When  the  point  on  the  rim  of  the  movmg  circle  is  at  P,  let  A  be 


the  point  of  the  circle  which  is  in  contact  with  the  line  EF,  and 
let  AB  be  the  diameter  of  the  circle  which  passes  through  A. 

At  the  instant  considered,  we  know  that  the  motion  of  the  point 
P  on  the  rim  of  the  circle  is  perpendicular  to  the  line  AP  (see 
example  1  on  p.  9).  Thus  since  APB  is  a  right  angle,  the  motion 
must  be  along  BP.    Thus  BP  is  the  tangent  to  the  cycloid. 

If  EF  is  supposed  horizontal,  the  angle  6  between  the  tan- 
gent at  P  and  the  horizontal  is  equal  to  the  angle  PAB,  so  that 


THE   CYCLOIDAL   PENDULUM  267 

the  radius  of  the  circle  throiigli  F  will  make  an  angle  2  6  with 
the  vertical. 

Suppose  that  the  circle  rolls  along  EF  until  the  tangent  to  the 
cycloid  at  F  makes  an  angle  6  -i-  dd  with  the  horizontal.  The 
radius  at  P  must  now  make  an  angle  2  (^  +  dd)  with  the  vertical, 
so  that  the  circle  must  have  rotated  through  an  angle  2  dd.  Since 
the  motion  of  F  may  be  regarded  as  one  of  rotation  about  A,  the 
small  element  of  path  ds  described  by  F  will  be  given  by 

ds  =  AF-2  dO. 


Now  AF  =  AB  cos  6  =  D  cos  6,  where  D  is  the  diameter  of  the 
rolling  circle.    Thus 

ds  =  21)  cos  e  dd, 

giving,  on  integration,        s  =  2  i>  sin  6. 

No  constant  of  integration  is  required  if  we  agree  to  measure  s 
from  the  pomt  at  which  ^  =  0,  i.e.  the  lowest  point  of  the  cycloid. 

The  property  of  the  cycloid  is  now  proved,  and  we  see  that 
equation  (99)  is  true  throughout  the  motion  of  a  point  which 
describes  a  cycloid,  this  being  generated  by  the  rolling  of  a  circle 
of  diameter  D  given  by 

21)  =  |,- 

212.  If  the  cycloid  is  given,  the  frequency  h  of  the  simple  har- 

\  (J  .       .    27r 

monic  motion  will  be  k  =  \  9^'  and  the  period  is  — —  >  or 


Thus  the  motion  is  of  the  same  period  as  that  of  a  simple 
pendulum  of  length  2D. 

213.  The  importance  of  cycloidal  motion  is  as  follows.  It  has 
been  seen  that  the  motion  of  a  simple  pendulum  is  only  strictly 
simple  harmonic  when  the  amplitude  is  so  small  that  it  may  be 
treated  as  infinitesimal.    For  finite  amplitudes  the  motion  is  not 


268  .MOTION  UNDER  A  VAlilABLE   FOECE 

simple  harinonic,  and  consequently  the  period  is  different  from  that 
of  the  simple  harmonic  motion  described  when  the  amplitude  is 
very  small.  Thus  the  period  depends  on  the  amplitude,  so  that  a 
clock  which  beats  true  seconds  when  the  pendulum  swings  through 
one  angle  will  gain  or  lose  as  soon  as  the  pendulum  is  made  to 
swing  through  any  different  angle.  A^ariations  of  amplitude  must 
always  occur  during  the  motion  of  any  pendidum,  and  these  cause 
irregularities  m  the  timekeeping  of  the  clock. 

AVe  have,  however,  seen  that  if  a  particle  describes  a  cycloid, 
the  period  is  independent  of  the  ampHtude,  so  that  variations  of 

amplitude  cannot  affect  the 
timekeeping  powers  of  a  par- 
ticle moving  in  a  cycloid. 

The  simplest  way  of  caus- 
ing a  particle  to  move  in  a 
cycloid  is,  in  practice,  to  sus- 
pend it  from  a  fixed  point  by 
"■~-~-._._  _  _^,---'"  a  string,  in  such  a  way  that 

during  its  motion  the  string 
wraps  and  unwraps  itself 
about  two  vertical  cheeks.  If  the  curve  of  these  cheeks  is  rightly 
chosen,  the  particle  can  be  made  to  describe  a  cycloid,  and  it  can 
easily  be  shown  that  the  curves  of  the  cheeks  must  be  portions  of 
two  cycloids  each  equal  to  the  cycloid  which  is  to  be  described  by 
the  particle. 

EXAMPLES 

1.  In  cycloidal  motion  prove  that  the  vertical  component  of  the  velocity  of 
the  particle  is  greatest  when  it  has  described  half  of  its  vertical  descent. 

2.  A  particle  oscillates  in  a  cycloid  under  gravity,  the  amplitude  of   the 

motion  being  b  and  the  period  being  t.    Show  that  its  velocity  at  a  time  t  meas- 

,  .                    .,.          .            .    2Trb    .     2irt 
ured  from  a  position  oi  rest  is sin 

T  T 

3.  A  particle  of  mass  m  slides  on  a  smooth  cycloid,  starting  from  the  cusp. 
Show  that  the  pressure  at  any  point  is  2  mg  cos  i//,  where  xp  is  the  inclination  to 
the  horizontal  of  the  direction  of  the  particle's  motion. 


MUTiUX  ABOUT  A  CENTER  OF  FORCE  269 

Motion  of  a  Pakticle  about  a  Center  of  Force 
Force  Projiortional  to  the  Distance 

214.  Let  us  suppose  that  a  particle  moves  under  no  forces  ex- 
cept an  attraction  to  a  fixed  point  0,  the  force  of  attraction  being 
directly  proportional  to  its  distance  from  0. 

Taking  0  as  origin,  let  the  coordinates  of  the  point  P,  the  posi- 
tion of  the  particle  at  any  instant,  be  denoted  by  x,  y,  z.  Let  the 
force  acting  on  the  particle  be  /x  •  OP  directed  along  PO,  where  /i 
is  a  constant.  The  components  of  this  force  along  the  three 
coordinate  axes  are 

^^,  ^y,  fXZ. 

The  components  of  acceleration  are,  as  in  §  177, 

d^x  dPy  d?z 

lie'         'de'         d^' 

Thus  the  equations  of  motion  of  the  particle  are 

d'^x 
^^  =  —/*«,  (100) 

m^^  =  -,.z.  (102) 

These  three  equations  are  all  of  the  same  t}-pe,  namely  the  type 
of  equation  which  denotes  simple  harmonic  motion.  Thus  the  foot 
of  the  perpendicular  from  the  moving  point  on  to  each  of  the 
coordinate  axes  moves  with  simple  harmonic  motion. 

The  solution  of  equation  (100)  has  already  been  seen  to  be 

x=A  cos 2}{t  —  e), 
where  p^=  fi/m.    This  can  be  written 

x=A  cos  pe  cos  jJt  +  A  sin  pe  sin  2>i, 
or  again  x  =  C  cos  pt  +  D  sin  j^t, 


270  MOTION   UNDER  A  VARIABLE  FORCE 

if  we  introduce  new  constants  C,  D  to  replace  the  constants 
^  cos^e  and  As>inpe.  The  other  two  equations  have  similar 
solutions,  so  that  we  can  take  the  complete  solution  to  be 

X  =  C  C0&  pt -{- DQiwjot,  (103) 

y  =  C  cos  pt  +  B' Bin  pt,  (104) 

z  =  C"  cos  pt  +  D"  sm  pt  (105) 

We  can  always  solve  the  equations 

C  +  rC'+sC"=0,  (106) 

.D  +  rI)'+sD"=0,  (107) 

and  so  obtain  values  of  r  and  ,s'  for  which  these  relations  are  true. 
Let  us  multiply  equations  (104)  and  (105)  by  these  values  of  r  and 
s,  and  add  corresponding  sides  to  the  corresponding  sides  of  equa- 
tion (103).    We  obtain 

{x  +  ri/-\-  sz)  ^{C+rC'+s  C")  cos  pt  +  (D  +  rD'  +  sD")  sin  j^t 

=  0,  (108) 

since  equations  (106)  and  (107)  are  satisfied.  The  meaning  of 
equation  (108)  is  that  for  all  values  of  t  we  have  the  relation 
X  +  ri/  +  s^  =  0,  and,  therefore,  that  throughout  its  motion  the 
particle  remains  in  the  plane  of  which  this  is  the  equation. 

The  axes  of  coordinates  have  been  supposed  to  be  chosen  arbi- 
trarily. We  can  always  choose  the  axes  so  that  the  plane  in  which 
the  whole  motion  takes  place  is  that  of  xy.  The  motion  is  then 
given  by  two  equations  of  the  form 

X  =  C  cos  pt+  D  sin  pt, 
y  =  C'  cos  pt  +  D'  sin  pt. 
Solving  for  sin^?^  and  cos  pt  we  obtain 

C'x  -  Cy 


sm.pt 


cos  jyt  =  — 


CD  -  CD' 
D'x—Dy 


CD  -  CD' 
so  that  on  squaring  and  adding,  we  obtain 

{C'x  --Cyf  -^  {D'x  - Dyf  ={CD  -  CD'f. 


MOTION  ABOUT  A  CENTER  OF  FORCE  271 

This  is  the  equation  of  an  ellipse. 

Thus  the  most  general  motion  possible  for  the  particle  consists 
in  describing  the  same  ellipse  over  and  over  again.  The  period  is 
2  tt/^,  this  being  the  time  required  for  cos  pt  and  sin  pt  both  to 
repeat  their  values. 

215.  The  axes  of  x,  y  are  still  undetermined ;  let  us  imagine 
them  to  be  the  principal  axes  of  the  ellipse. 

Then  if  we  suppose  the  time  measured  from  one  of  the  instants 
at  which  the  particle  is  at  one  of  the  extremities  of  the  major 
axis,  we  shall  have  equations  of  the  form 

X  =  A  cos  pt, 

y  =  B  sin  pt. 

Thus  pt  is  the  eccentric  angle  of  the  ellipse  described  by  the 
particle,  so  that  the  eccentric  angle  increases  with  uniform  angular 

velocity  p  or  ■\\ —    The  motion  repeats  itself  as  soon  as  p  mcreases 

y'ln  I —  j — 

by  2  TT.  Thus  the  frequency  is  p  or  \^  — '  while  the  period  is  2  tt  \\~- 

216.  This  motion  is  realized  experimentally  in  the  motion  of 
a  pendulum  which  is  not  constrained  to  move  in  one  vertical 
plane,  but  of  which  the   deviations   from  the  vertical 
remain  small. 

Let  the  pendulum  be  of  length  a,  and  let  its  bob  be 
displaced  from  its  equilibrium  position  0  to  some  near 
point  P,  such  that  the  angle  PCO  may  be  treated  as 
small.  Calling  tliis  angle  6,  the  weight  of  the  bob  may 
be  resolved  into  mg  cos  6  along  CP,  which  is  exactly  neu- 
tralized by  the  tension  of  the  strmg,  and  a  force  mg  sin  6  o 
along  PO.    li  6  is  small,  sin  0  may  be  put  equal  to  6,  and     Fig-  i^-* 

OP 

this  in  turn  to Thus  the  bob  mav  be  supposed  to  experience 

a 

a  force  —  OP  along   OP.    The   motion  is  therefore  of   the   kind 
a  ^ 

which  has  been  described,  the  value  of  /x  bemg  — ^ »  and  the  value 

r  ^ 

of  p  therefore  beiug   yj-  ■    Thus  we  see  that  a  hanging  weight 


pA 


272  :\1()T10K   UNDEli  A  VAKIABLE  FOllCE 

drawn  from  its  position  of  equilil)rium  and  projected  in  any  way 

will  always  describe  an  ellipse  in  the  horizontal  plane  in  which  it 

is  free  to  move,  having  the  point  immediately  below  its  point  of 

suspension  as  center. 

An  arrangement  may  sometimes  be  seen  at  village  fairs  in  England,  in 
which  the  showman  ingeniously  takes  advantage  of  this  result.  A  weight 
is  suspended  by  a  string,  and  a  skittle  is  placed  on  the  floor  exactly  under 
the  point  of  suspension  of  the  weight.  Passers-by  are  invited  to  pay  an 
entrance  fee  and  compete  for  a  prize  which  is  awarded  to  any  one  who  can 
project  the  weight  so  that  on  its  return  it  knocks  the  skittle  over.  The 
problem  is,  of  course,  as  impossible  as  that  of  describing  an  ellipse  which 
shall  pass  through  its  own  center. 

217.  Another  way  in  which  motion  under  a  force  proportional 
to  the  direct  distance  may  be  realized,  is  as  follows :  An  elastic 
string  of  natural  length  I  has  one  end  fastened  to  a  small  particle 
which  is  free  to  move  on  a  smooth  horizontal  table ;  tire  other  end, 
after  passing  through  a  small  hole  in  the  table,  is  fastened  to  a 
fixed  point  at  a  distance  I  from  the  hole.  If  the  particle  is  pulled 
away  from  the  hole  to  a  point  distant  r  from  it,  the  total  length 

T 

of  the  string  is  /  +  r,  so  that  its   tension  is  -  X,  where  A,  is  its 

modulus  of  elasticity.  The  fdrce  acting  on  the  particle,  namely 
the  tension  of  the  string,  is  tlrerefore  proportional  to  the  distance 
of  the  particle  from  a  fixed  point  —  namely  the  hole  in  the  table 
—  and  its  direction  is  toward  the  hole.  Thus  the  particle  will 
move  in  elliptic  motion  on  the  table. 

EXAMPLES 

i.  The  point  P  is  describing  an  ellipse  under  an  attractive  force  to  the  center, 
and  p  is  the  corresponding  point  on  the  auxiliary  circle.  Show  that  p  moves 
round  the  auxiliary  circle  with  uniform  velocity. 

2.  A  particle  describes  an  ellipse  about  a  center  of  force,  the  attraction 
being  that  of  the  direct  distance.  Show  that  the  radius  vector  from  the  center 
of  the  ellipse  to  the  particle  sweeps  out  equal  areas  in  equal  times. 

3.  A  particle  is  describing  an  ellipse  under  a  force  proportional  to  the  dis- 
tance, when  it  receives  a  blow  in  a  direction  parallel  to  the  major  axLs  of  the 
ellipse.  Show  that  the  minor  axis  of  the  new  orbit  is  the  same  as  that  of  the 
old,  and  show  how  to  find  tlie  change  produced  in  the  major  axis. 


MOTION  ABOUT  A  CENTER  OF  FORCE  273 

4.  A  particle  is  acted  on  by  attractions  to  a  number  of  centers  of  force,  each 
being  proportional  to  the  distance.    Show  that  it  describes  an  ellipse. 

low  could  a  mechanical  model  be  constructed  to  illustrate  this  motion  ? 

^<<A  particle  is  acted  on  by  a  repulsion  proportional  to  its  distance  from  a 
center  of  force.    Show  that  it  describes  a  hyperbola. 

6r^how  that  in  the  last  question  the  radius  vector  joining  the  particle  to 
the  center  of  force  sweeps  out  equal  areas  in  equal  times. 


General  Theory  of  Motion  about  a  Center  of  Force 

218.  Suppose  that  we  have  a  particle  acted  on  only  by  a  force 
directed  towards  a  fixed  center  of  force,  the  magnitude  of  this  force 
being  any  function  of  the  distance  from  the  center.    . 

Let  0  be  the  center  of  force,  P  the  position  of  the  particle  at 
any  instant,  and  PP'  the  direction  of  the  velocity  of  the  particle 
at  this  instant.  Then 
the  plane  OPP'  con- 
tains the  velocity  of  the 

particle,  which  is  along    (     ■  0« 

PP',  and  also  the  accel- 
eration, which  is  along 
PO.  Hence,  after  any 
short  interval  the  veloc- 
ity of  the  particle  will  still  be  in  the  plane  OPP'.  The  particle  is 
still  in  this  plane,  say  at  P' ,  so  that  the  acceleration  which  is 
along  P'  0  is  also  in  this  plane. 

Hence  we  can  show  that,  after  a  further  small  interval,  the  posi- 
tion, velocity,  and  acceleration  of  the  particle  are  all  in  the  plane 
OPP' ,  and  so  we  can  proceed  indefinitely. 

It  follows  that  the  particle  will  never  leave  the  plane  OPP". 
We  accordingly  have  the  theorem : 

The  orhit  described  hy  a  particle  ciboiit  a  fixed  center  of  force  lies 
entirchj  in  one  jjlctne. 

Tliis  theorem  has  already  been  exemplified  in  §  214  by  the 
case  of  the  orbit  described  under  an  attraction  proportional  to  the 
distance  from  the  center  of  force. 


274  MOTION   UNDEK  A  VAIUAP.LE   FORCE 

Moment  of  a  Velocity 

219.  The  velocity  of  a  point  is  a  vector,  aud  the  line  of  action 
of  this  vector  may  be  supposed  to  be  the  line  through  the  moving 
point  in  the  direction  of  its  velocity.  We  can  define  the  moment 
of  a  velocity  in  just  the  same  way  as  the  moment  of  a  force  has 
been  defined.  Moreover,  all  the  properties  of  the  moments  of  a 
force  followed  from  the  fact  that  forces  could    be   compounded 

according  to  the  parallelogram  law,  so 
that  the  same  properties  will  be  true  of 
the  moments  of  velocities,  because  veloci- 
ties also  can  be  compounded  according  to 
the  parallelogram  law. 

Let  P  be  a  particle  descri1)ing  an  orbit 
about  0,  and  let  0()  be  a  perpendicular 
from  0  on  to  the  line  through  P  m  the 
direction  of  the  particle's  velocity.    Then  the  moment  of  the  par- 
ticle's velocity  about  0  is  0()  X  (velocity  of  particle). 

After  a  short  interval  dt,  let  the  particle  be  at  P'.  Its  velocity 
at  P'  is  compounded  of  its  velocity  at  P  together  with  dt  times  its 
acceleration  at  P.    Hence 

(moment  about  0  of  velocity  at  P') 

=  (moment  about  O  of  velocity  at  P) 

+  (moment  about  0  of  \dt  x  acceleration  at  P] ). 

The  acceleration  at  P  being  along  PO,  the  last  term  of  this 
equation  is  zero,  so  that  we  see  that  the  moments  about  O  of  the 
velocities  at  P  and  at  P'  are  equal. 

We  can  extend  this  step  by  step  as  in  the  former  theorem,  and 
obtain  finally  that 

The  moment  ahout  0  of  the  velocity  of  a  particle  describing  an 
orbit  about  0  is  constant. 

220.  We  have  supposed  that  the  particle  moves  from  P  to  P' 
in  time  dt,  so  that  if  v  is  its  velocity  at  P,  then  PP'  =  v  dt.  As 
the  particle  describes  its  orbit,  the  line  OP  may  be  regarded  as 


MOTION  ABOUT  A  CENTER  OF  FORCE  275 

sweeping  out  an  area  in  the  plane  of  the  orbit.    The  area  described 
in  time  dt  is  the  small  triangle  OPP'.    We  now  have 

area  described  in  time  dt 
=  area  OPP' 
=  \OQ-PP' 
=  \0Q  -vdt 
=  '^dt  X  moment  of  velocity  about  0. 

Thus  the  area  described  per  unit  time  is  half  the  moment  of  the 
velocity  about  0,  and  this  by  the  theorem  of  the  last  section  is  a 
constant.    Thus  we  have  the  theorem : 

Equal  areas  are  described  in  equal  times. 

Differential  Equation  of  Orbit 

221.  The  theorem  just  proved,  in  combination  with  the  theo- 
rem of  the  conservation  of  energy,  enables  us  to  determine  the 
equation  of  the  orbit  in  which  a  parti- 
cle will  move.  This  equation  is  most 
conveniently  expressed  in  polar  coordi- 
nates, the  center  of  force  being  taken 
as  origin. 

If  r,  6  are  the  polar  coordinates  of  the 
particle,  the  velocity  may  be  regarded  ^^^-  ^^^ 

as  compounded  of  a  velocity  —  along  OP,  and  a  velocity  r  — 
at  right  angles  to  OP. 

The  velocity  is  therefore  given  by 

The  moment  of  the  velocity  about  0  is  equal  to  the  moment  of 
the  second  component,  for  the  moment  of  the  first  com[)onent 

de 

vanishes.    Thus  the  moment  of  the  velocitv  about  0  is  ;•  X  r  — -  > 

,    .  ,  .    -  -  ,    "^     -  dt 

and  smce  this  lias  a  constant  value,  say  h,  we  nave 

r'—=h.  (100) 

dt 


276 


.MOTION   UNDER  A  VARIABLE  FORCE 


If  m  is  the  mass  of  the  particle,  and  if /(r)  is  tlie  attraction  per 

unit  mass  when  at  a  distance  r  from  0,  we  find  that  the  potential 

energy  of  the  particle  is  ^r 

m  I  f{r)dr. 

t/x- 


The  kinetic  energy  is  ^  mv^,  or 


dt  \dt 


Expressing  that  the  total  energy  is  constant,  we  have 


(110) 


where  ^  is  a  constant. 

Equations  (109)  and  (110)  lead  to  the  differential  equation  of 
the  orbit.    We  have,  since  r  and  6  are  both  fimctions  of  t, 

dr  _  dr  dd 
di'~dd  dt' 

so  that  equation  (110)  may  be  ex]3ressed  in  the  form 


+  7^ 


dd 

di 
dr\- 


+ 


'f/' 


r)  dr  =  E, 


and  on  eliminating  —  from  this  and  equation  (109),  we  have 
dt 


16 


+  r 


ir- 


-+2      f(r)dr=E, 


(111) 


the  differential  equation  of  the  orbit. 


Law  of  Inverse  Square 

222.  Let  us  now  suppose  that  the  attraction  follows  the  law  of 
the  inverse  square  of  the  distance,  so  that 


fir)  = 


where  /a  is  a  constant.    Then 


X 


f(r)dr^-^, 
r 


(112) 


LAW  OF  INVERSE   SQUAKE  277 

and  equation  (111)  becomes 


cW 


whence  we  obtain        clQ  = 


giving,  on  integration,    a  =  sin~  ^  — j^=r^^^  +  e, 


4 


where  e  is  a  constant  of  integration. 
Simplifying,  this  becomes 

h      IX 


r 


._-^vv:7 


and  if  we  compare  with  the  equation 

-  —  1  =  e  cos  6, 
r 

we  see  that  equation  (113)  represents  a  conic,  having  the  origin 

as  focus,  and  being  of  semi-latus  rectum  /  =  —  and  eccentricity 

I 'Eh'  ^ 

e  =  -\\\  -\ —  •    In  order  that  the  line  ^  =  0  may  coincide  with 

the  major  axis  of  the  conic,  the  vakie  of  e  must  be . 

223.  We  notice  that  if 

E  is  positive,  then  e  >  1,  and  the  orbit  is  a  hyperbola ; 
E  is  zero,         then  e  =  1,  and  the  orbit  is  a  parabola; 
E  is  negative,  then  c  <  1,  and  the  orbit  is  an  elhpse. 

Thus  the  class  of  conic  described  depends  solely  on  the  value 
of  E,  and  not  on  that  of  li.  And  it  should  be  noticed  that,  if  we 
are  given  the  point  of  projection  of  a  particle,  and  also  its  velocity 
of  projection,  the  value  of  E  is  determined,  for  by  equation  (110) 


E  =  V 

r 


278  MOTION  UNDER  A  VAIUABLE  FORCE 

Thus  the  class  of  conic  described  depends  only  on  the  velocity 
of  projection,  and  not  on  the  direction :  the  conic  is  a  hyperbola, 
parabola,  or  ellipse,  according  as 

v  >  =  or  <  — ^  • 
r 

The  actual  eccentricity  depends  on  both  E  and  h,  for  if  e  is  the 
eccentricity,  we  have  ,  2 


224.  In  order  that  thfi-parttcle  may  describe  a  circle  we  must 
have  e'—O,  and  therefore  „, , 

1+^  =  0. 

Putting  E  =  v^ and  h  =  pv  (so  that  p  is  the  perpendicular 

r 

from  the  center  of  force  on  to  the  direction  of  projection),  this 
reduces  to  29 

p         r 

Since  p  is  necessarily  less  than  r,  neither  term  in  this  equation 
can  be  negative.  Thus,  in  order  that  the  equation  may  be  satisfied, 
both  terms  must  vanish,  and  we  must  have 

p  =  r  and  v^  —  —• 
r 

The  first  equation  expresses  that  the  projection  must  be  at  right 
angles  to  the  line  joining  the  particle  to  the  center  of  force.  The 
second  equation,  w^hich  can  be  written 

^2       ^, 

shows  that  the  attractive  force  must  just  produce  the  acceleration 
appropriate  to  motion  in  a  circle  of  radius  r. 


LAW  OF  INVERSE   SQUARE  279 

225.  For  an  elliptic  orbit,  the  periodic  time  is  that  required  to 
sweep  out  an  area  irab,  where  a,  h  are  the  semi-axes  of  the  ellipse. 
Since  the  area  is  swept  out  at  a  rate  ly  h  per  unit  time,  the  periodic 
time  T  will  be  , 

TTOh 

The  semi  latus-rectum  I  is  equal  to  —  >  and  has  also  been  seen 
to  be  equal  to  —  >  so  that 


.2  Ct 


h  —  y/al  =  h  \  -  > 
Mm 


,  ^2  irah       2  ira^  /i  i  ^, 

whence  T  =  — ; —  =  — ^=-  •  (114) 

li  VA'' 

Since  this  does  not  depend  on  the  eccentricity,  it  is  clear  that 
the  periodic  time  of  any  orbit  is  the  same  as  that  in  a  circle  of 
radius  e([ual  to  the  semi  major-axis. 

226.  The  law  of  force  of  the  inverse  square  is  that  of  gi-avitation  : 
the  law  which  we  have  been  investigating  is  therefore  that  which 
governs  the  motions  of  the  planets  in  their  orbits  round  the  sun, 
as  well  as  the  motions  of  comets  and  meteorites.  For  reasons 
Avhich  cannot  be  explained  here,  the  conies  described  by  the  planets 
are  all  of  them  ellipses  of  small  eccentricity.  A  wider  range  is 
found  in  the  orbits  of  comets.  These  bodies  generally  come  from 
far  outside  the  solar  system.  To  a  close  approximation  many  of 
them  may  be  treated  as  coming  from  infinity,  and  as  starting  with 
relatively  small  velocity.  In  tliis  case  the  orbit  is  approximately 
parabolic. 

Kepler's  Laws 

227.  Long  before  the  theory  of  the  planetary  orbits  had  been 
worked  out  mathematically  by  Newton,  three  of  the  principal 
laws  governing  the  motion  of  the  planets  had  been  discovered 
empirically  by  Kepler.    Kepler's  three  laws  are  as  follows : 

Law  I.  Every  planet  describes  an  ellipse  having  the  sun  in  one 
of  its  foci. 


280 


MOTION   UNDER  A  VAKIABLE  FORCE 


V 


Law  II.  2'he  areas  described  by  the  radii  draivn  from  the  planet 
to  the  sun  are,  in  the  same  orbit,  proportional  to  the  times  of 
describing  them. 

Law  III.  The  squares  of  the  periodic  times  of  the  various  orbits 
are  proportional  to  the  cubes  of  their  major  axes. 

From  the  first  of  these  laws  Newton  proved  that  the  law  of 
force  between  the  planets  and  the  sun  must  be  the  law  of  tlie 
mverse  scjuare.  The  third  law  is  seen  to  express  the  same  fact  as 
equation  (114). 

Motion  of  Two  Particles  about  One  Another 

228.  A  pair  of  objects  known  as  a  double  star  is  of  common 
occurrence  in  the  sky.  This  consists  of  two  stars  describing  orbits 
about  one  another,  neither  star  being  fixed. 

By  the  theorems  proved  in  Chapter  IX,  the  center  of  gravity  of 
the  two  stars  must  either  remain  at  rest,  or  else  must  move  with 
uniform  velocity  in  a  straight  hne,  in  which  case  it  may,  as  we 
have  seen,  be  treated  as  fixed,  provided  all  motion  is  measured 
relative  to  a  frame  of  reference  moving  with  it. 

Let  A,  B  be  the  positions  of  the  two  stars  at  any  instant,  and  let 
G  be  their  center  of  gravity.  Let  the  masses  of  the  stars  be  m,  m', 
and  let  a,  b  denote  their  distances  from  G.    Then 

m  _m'  _  m  -{-  m' 
b        a         a  +  b 


(115) 


The  complete  law  of  gravitation  is  expressed  by  the  law 


r^ 

where  m,  m'  are  the  two  masses,  r  the  distance  between  them,  7  a 

constant  whose  value  can  be  found  by  experiment,  and  F  is  the 

force  of  attraction  between  the  two  masses.    Thus  the  force  acting 

on  the  star  B  is  , 

^            mm 
F=  7 -1 

^  («  +  bf 


LAW  OF  INVERSE   SQUARE  281 

acting  along  BA.  This  force  can  always  be  regarded  as  acting  from 
the  fixed  point  G,  for  its  line  of  action  is  always  BG.  Moreover, 
its  magnitude  per  unit  mass  of  star  B  is 


(a  +  ly 

or,  from  relations  (115), r-r-,' 

^       '        {m  +  m'f  V 

This  is  a  force  —  acting  towards  G  if  we  take 
r 

(ill  +  Til) 

Thus  the  two  stars  each  describe  a  conic  about  the  center  of 
gravity  of  the  two.  It  is  possible  astronomically  to  observe  the 
values  of  the  periodic  time  T  and  the  major  axes  of  the  orbits  of 
these  conies.  From  these  quantities  we  can  determine  the  values 
of  /A,  so  that  we  know  the  values  of 


(ill  +  rii'Y' '  {ill  +  m'f ' 

and  these  at  once  lead  to  the  values  of  m,  m'.    In  this  way  it  has 
been  found  possible  to  determine  the  masses  of  some  of  the  stars. 

EXAMPLES 

(Take  the  gravitation  constant  to  be  y  =  G6.6  x  10"  *  in  centimeter-gramme-second  units.) 

1.  Given  that  the  earth  attracts  as  though  its  mass  were  concentrated  at  its 
center,  and  that  the  value  of  g  at  tlie  equator,  distant  0.378  x  10^  centimeters  from 
tlie  earth's  center,  is  978. 1  centimeters  i^er  second  i^er  second,  find  tlie  mass  of  the 
eartli. 

2.  Taking  the  masses  of  tlie  earth  and  moon  as  6.14  x  10-"  and  7.94  x  lO^* 
grammes  respectively,  and  assuming  their  distance  apart  to  be  always  3. 84  x  lOi** 
centimeters,  find  the  periodic  time  of  the  moon. 

3.  Taking  the  siui's  mass  to  be  2  x  lO^^  grammes,  and  the  year  to  be  365.24 
days,  find  the  semi  major-axis  of  the  earth's  orbit,  regarding  the  sun  as  a  fixed 
center  of  force. 

4.  If  the  sun's  mass  is  324,000  times  that  of  the  earth,  by  how  much  must 
the  result  of  question  3  be  altered  when  the  sun's  motion  is  taken  into  account  ? 


282  MOTION   UNDER  A  VAKIABLE   FORCE 

5.  Taking  the  mass  of  Jupiter  to  be  -jo^^q  of  the  mass  of  the  sun,  and  its 
greatest  distance  from  the  sun  to  be  498J  million  miles,  show  that,  on  account  of 
Jupiter's  attraction,  the  sun  will  describe  an  ellipse  of  semi  major-axis  ecjual  to 
about  4(51,000  miles,  and  find  the  length  of  Jupiter's  year. 

6.  The  maximum  velocity  attained  by  the  earth  in  its  orbit  is  3,000,000 
centimeters  per  second,  and  the  minimum  velocity  is  2,920,000  centimetere  per 
second.    Find  the  eccentricity  of  the  earth's  orbit. 


GENERAL  EXAMPLES 

1.  A  particle,  attached  by  a  string  to  a  point,  has  just  sufficient  energy 
to  make  complete  revolutions  in  a  vertical  circle.  Show  that  the  tension 
of  the  string  is  zero  and  six  times  the  weight  of  the  jsarticle  respectively, 
when  the  particle  is  at  the  highest  and  lowest  points  of  its  path. 

2.  A  particle  moves  under  gravity  in  a  vertical  circle,  sliding  down  the 
convex  side  of  a  smooth  circular  arc.  If  its  velocity  is  that  due  to  a  height 
h  above  the  center,  show  that  it  will  fly  off  the  circle  when  at  a  height  f  h 
above  the  center. 

3.  If  the  angle  a  through  which  a  simple  pendulum  swings  on  each 
side  of  the  vertical  is  small,  but  not  infinitesimal,  show  that  to  a  first 
approximation  the  time  of  oscillation  is 


Deduce  that  a  pendulum,  which  beats  seconds  accurately  when  j^erforming 
infinitesimal  oscillations,  would  lose  about  40  seconds  a  day  when  attached 
to  a  clock  which  caused  it  to  oscillate  to  5  degrees  on  each  side  of  the 
vertical. 

4.  A  train  is  moving  uniformly  round  a  curve  at  60  miles  an  hour,  and 
in  one  of  the  carriages  a  seconds  pendulum  is  foimd  to  beat  121  times 
in  two  minutes.  Show  that  the  radius  of  the  curve  is  about  a  quarter 
of  a  mile. 

5.  One  end  of  an  elastic  string,  natural  length  o,  modulus  X,  is  tied  to 
a  fixed  point  on  a  smooth  horizontal  table,  and  the  other  end  is  tied  to  a 
particle  of  mass  m  which  rests  on  the  table.  If  the  mass  is  pulled  to  a 
distance  2  a  from  the  other  end  of  the  string,  and  is  then  let  go,  show  that 

it  will  return  to  its  original  position  at  regular  intervals  2(7r  +  2)-%/ 

6.  Two  balls  weighing  W^  and  TF.,  i")Ounds  are  connected  by  a  thread  a 
feet  long  ;  and  W^  is  held  in  the  hand  while  W^  is  whirled  round.  Deter- 
mine the  motion  which  ensues  if   W,   is  released  from  rest  when    W„ 


EXAMPLES  283 

is  moving  with  velocity  V  at  iiu;liuatiou  a  ;    and  prove  that  in  the  air 
the  tension  of  the  thread  is 

7.  Two  masses,  rn^  and  m., ,  are  connected  by  a  weightless  spring  of  such 
strength  that  when  m^  is  held  fixed,  7n^  performs  n  vibrations  a  second. 
Show  that  if  //;.,  ^e  held  fixed,  m^  will  perform  n  ^mjm.^^  vibrations  a  sec- 

VfJl       -I-    111 
— 5  vibra- 

tions  per  second,   the  vibrations  in  every  case  being  in  the  line  of  the 
spring. 

8.  A  particle  of  mass  m,  moving  in  a  smooth  curved  tube  of  any  shape, 
is  in  equilibrium  under  the  tensions  of  two  elastic  strings  in  the  tube,  of 
natural  lengths  I,  V  and  moduli  of  elasticity  X,  X',  of  which  the  other  ends 
are  attached  to  fixed  points  of  the  tube.  If  the  particle  makes  oscillations, 
large  or  small,  in  the  tube,  show  that  the  time  of  oscillation  is 


9.  A  string  passes  through  a  small  hole  in  a  smooth  horizontal  table, 
and  has  equal  particles  attached  to  its  ends,  one  hanging  vertically  and  the 
other  lying  on  the  table  at  a  distance  a  from  the  hole.  The  latter  is  pro- 
jected with  a  velocity  V^  perpendicular  to  the  string.  Show  that  the 
hanging  particle  will  remain  at  rest,  and  that  if  it  be  slightlv  disturbed, 

10.  A  particle  moves  in  a  circular  groove,  imder  an  attraction  ~  to  a 

point  P  which  is  in  the  plane  of  the  circle  and  distant  h  from  its  center. 
The  particle  is  projected  with  velocity  V  from  the  point  of  the  circle  nearest 
to  P.    Show  that  for  the  particle  to  perform  complete  revolutions,  the  value 

4yuft 


the  time  of  a  small  oscillation  will  be 


of  V^  must  not  be  less  than 


a^  -  Ifl 


11.  A  smooth  ellipse,  semi-axes  a  and  h,  is  placed  with  its  major  axis 

vertical,  and  a  particle  is  projected  along  the  concave  side  of  the  arc,  with 

velocity  due  to  a  height  h  above  the  center.    Find  the  point  at  which  the 

particle  will  leave  the  arc,  and  show  that  it  will  pass  through  the  center  of 

the  ellipse  if 

^      8  a2  +  jfi 

h  = — • 

6aV3 


284  MOTION  UNDER  A  VARIABLE  FORCE 

12.  A  particle  is  constrained  to  move  in  a  circle  of  radius  a,  under  an 
attraction  fir  per  unit  mass  to  a  point  inside  the  circle  distant  c  from  its 
center.  If  the  particle  be  placed  at  its  greatest  distance  from  this  point, 
and  started  with  an  infinitesimal  velocity,  i)rove  that  it  will  pass  over  the 
second  quadrant  of  the  circle  iu  a  time 


\i 


-  log(V2  +  l). 

flC 

13.  A  particle  describes  an  ellipse  about  a  center  of  force  in  one  focus. 
Show  that  the  velocity  at  the  end  of  the  minor  axis  is  a  mean  proportional 
between  the  velocities  at  the  ends  of  any  diameter. 

14.  A  comet  describes  a  parabola.  Show  that  its  velocity  perpendicular 
to  the  axis  of  its  orbit  varies  inversely  as  the  radius  vector  from  the  sun. 

15.  A  comet  of  mass  m,  describing  a  parabola  about  the  sun,  collides 
with  an  equal  mass  m  at  rest,  and  the  masses  move  on  together.  Show  that 
their  center  of  gravity  will  describe  a  circle  about  the  sun  as  center. 

16.  Assimiing  that  a  projectile,  after  allowing  for  variations  in  gravity, 
describes  an  ellipse  about  the  earth's  center  as  focus,  show  that  the  maxi- 
mum range  on  a  horizontal  plane  through  the  point  of  projection,  for  a 
given  velocity  i^,  is  ^  ^^,2/32 

where  R  is  the  distance  from  the  earth's  center  to  the  point  of  projection. 

17.  When  the  earth  is  at  the  end  of  the  major  axis  of  its  orbit,  a  small 

meteor,  of  mass  one  mih.  of  that  of  the  sun,  suddenly  falls  into  the  sun. 

2 
Show  that  the  length  of  the  year  will  be  diminished  bv  —  of  itself. 

'    m 

18.  A  planet  P  moving  about  the  sim  ^  picks  uji  a  small  meteor,  and 
consequently  has  its  velocity  reduced  by  one  nth  of  its  former  amount, 
although  unaltered  in  direction.  Treating  n  as  small,  show  that  the  eccen- 
tricity of  the  planet's  orbit  will  be  reduced  by  2  n  (e  +  cos  ^),  where  0  is  the 

angle  between  SP  and  the  major  axis  of  the  orbit. 

o,           ,         ,          ,                      .           ..-,,,                   ,     2n  sm^      .,, 
Show  also  that  the  new  major  axis  will  make  an  angle witn 

the  old  axes. 

19.  A  particle  describes  an  ellipse  about  the  focus.  Show  that  the 
greatest  and  least  angular  velocities  occur  at  the  ends  of  the  major  axis,  and 
also  that  if  a,  /3  be  these  angular  velocities,  the  mean  angular  velocity  is 


EXAMPLES  285 

20.  A  comet  describes  a  parabola  a])out  the  sun,  its  nearest  distance 
from  the  sun  being  one  third  of  the  radius  of  the  earth's  orbit,  supposed 
circular.  For  liow  many  days  will  the  comet  remain  within  the  earth's 
orbit  ? 

21.  If  the  attraction  on  a  particle  varies  as  the  inverse  square  of  the 
distance  from  a  center  of  force  O,  show  that  there  are  two  directions  in 
which  a  particle  can  be  projected  from  a  given  point  P  so  that  its  orbit 
may  have  a  given  major  axis.  If  OP  =  c,  and  if  atj  ,  a^  are  the  angles 
which  the  two  directions  of  projection  make  with  OP,  show  that 

cot  a^  cot  o'a  = 1> 


a 


where  a  is  the  semi  major-axis. 


22.  A  particle  is  projected  from  a  point  P  under  a  force  to  a  fixed  point 
^S  at  a  distance  R  from  P,  so  as  to  describe  a  circle  passing  through  S.  The 
initial  velocity  is  F,  and  the  moment  of  the  velocity  about  S  is  h.  Show 
that  the  particle  will  describe  a  semicircle  in  time 

4  A3  ^  ' 

23.  A  block  of  mass  M,  whose  upper  and  lower  faces  are  smooth  horizon- 
tal planes,  is  free  to  move  along  a  groove  in  a  parallel  plane,  and  a  particle 
of  mass  m  is  attached  to  a  fixed  point  in  the  upper  face  by  an  elastic  string 
whose  natural  length  is  a  and  modulus  X.  If  the  system  starts  from  rest 
"with  the  particle  on  the  upper  face,  and  the  string  stretched  parallel  to 
the  gi'oove  to  1  -I-  n  times  its  natural  length,  prove  that  the  block  will  per- 
form oscillations  of  amplitude 

(n  -I-  \)ain 
M  +  m 


and  period 


hU 


nMm 
\(M  +  m)' 


CHAPTEE  XI 
MOTION  OF  RIGID  BODIES 

229.,  The  present  chapter  is  devoted  to  a  discussion  of  the 
motion  of  rigid  bodies,  when  the  motion  is  such  that  the  bodies 
may  not  be  treated  as  particles. 

It  has  ah'eady  been  proved  in  §  66  that  the  most  general  motion 
possible  for  a  rigid  body  is  one  compounded  of  a  motion  of  trans- 
lation and  a  motion  of  rotation.  As  a  preliminary  to  discussmg 
the  general  motion  of  a  rigid  body  under  the  action  of  forces  of 
any  description,  we  shall  examine  in  greater  detail  than  has  so  far 
been  done  the  properties  of  a  motion  of  rotation. 

Angular  Velocity 

230.  We  have  seen  (§67)  that  for  every  motion  of  a  rigid  body 
in  which  a  point  P  remains  fixed,  there  is  an  axis  of  rotation, 
which  is  a  line  passing  through  P,  of  which  every  point  remains 
fixed.  If  a  rigid  body  is  moving  continuously  we  may  analyze  its 
motion  in  the  following  way.  We  select  a  definite  particle  P  of 
the  rigid  body,  and  we  refer  the  motion  to  a  frame  of  reference 
having  P  as  origin,  and  moving  so  as  always  to  remain  parallel 
to  its  original  position.  Relative  to  this  frame,  the  motion  of  the 
body  between  any  two  instants  is  a  motion  of  rotation  about  P. 

Now  let  the  two  instants  be  taken  very  close  to  one  another,- 
the  interval  between  them  being  dt.  Let  us  find  the  axis  of  rota- 
tion of  the  motion  wliich  takes  place  in  the  interval  dt,  and  call  it 
PQ.  Then  P(>  is  called  the  axis  of  rotation  at  the  instant  at  which 
the  interval  dt  is  taken. 

Let  us  suppose  that  during  the  interval  dt  the  rotation  of  the 
body  about   its  axis   of   rotation  PQ  is  found  to  be  a  rotation 

286 


ANGULAR  A^ELOCITY  287 

through  an  angle  (W.    Then  the  limit,  when  dt  is  made  to  vanish, 

of  the  rate  —  is  called  the   angular  velocity  of  the  body,  —  it 

measures  the  angle  turned  through  per  unit  time. 

Tlius  to  have  a  full  knowledge  of  the  motion  of  a  rigid  body  at 
any  instant  we  must  know 

(a)  the  direction  and  magnitude  of  the  velocity  of  the  point  P 
which  has  been  selected  to  give  a  frame  of  reference ; 

(b)  the  direction  of  the  axis  of  rotation  through  P ; 

(c)  the  magnitude  of  the  angular  velocity  about  the  axis  of 
rotation, 

231.  The  angular  velocity  has  associated  with  it  a  direction  — 
the  axis  of  rotation  —  and  a  magnitude.  Thus  it  may  be  repre- 
sented by  a  line.  We  shall  now  prove  that  it  is  a  vector,  i.e. 
that  angular  velocities  may  be  compounded  according  to  the 
parallelogram  law. 

Let  a  rigid  body  have  a  ro- 
tation about  P  compounded 
of  («)  a  rotation  of  angular 
velocity  «  about  an  axis  PQ, 
and  (b)  a  rotation  of  angular 
velocity  cd'  about  a  second 
axis  PQ'.  Let  the  lengths 
PQ,  PQ'  be  taken  proportional  to  co,  &)',  so  that  the  lines  PQ,  PQ' 
will  represent  the  directions  and  magnitudes  of  the  angular 
velocities  on  the  same  scale. 

Let  the  parallelogram  PQEQ'  be  completed,  and  let  L  be  any 
point  on  the  diagonal  PR.  Let  LX,  LN'  be  drawn  perpendicular 
to  PQ,  PQ'  respectively. 

In  time  dt  there  is,  from  the  first  angular  velocity,  a  rotation  of 
the  rigid  body  through  an  angle  (o  dt  about  PQ.  The  effect  of  tliis 
rotation  is  to  move  the  particle  of  the  body  wliich  originally  coin- 
cided with  L  through  a  distance  LN  •  (odt  at  right  angles  to  the 
plane  PLN.  Similarly  the  effect  of  the  rotation  about  PQ'  is  to 
move  the  same  particle  through  a  distance  LN'  •  (o'dt  at  right 


288  MOTION   OF  IIIGID   BODIES 

angles  to  the  plane  but  in  the  direction  opposite  to  that  of  the 
former  motion.    Thus  the  total  displacement  of  the  particle  is 

X^Yft)  dt  -  LN'  w'  dt.  (116) 

Since  L  is  on  the  diagonal  of  the  parallelogram,  we  see  that  the 
area  of  the  triangle  PLQ  is  equal  to  that  of  the  triangle  PLQ',  so 

^"^^^^  LNPQ  =  LN'  ■  FQ'. 

Again,  since  PQ,  PQ'  are  in  the  ratio  of  co :  &>',  this  equation  may 

be  written  in  the  form 

LN(o  =  LN'(o\ 

and  on  comparing  with  expression  (116)  we  see  that  the  displace- 
ment of  the  particle  L  vanishes. 

Thus  the  resultant  of  the  two  angular  velocities  is  a  motion  such 
that  the  points  P  and  L  Ijoth  remain  at  rest.  It  is  therefore  an 
angular  velocity  having  PP,  the  diagonal  of  the  parallelogram,  as 
axis  of  rotation. 

We  must  next  find  the  magnitude  of  this  angular  velocity. 
Let  us  denote  it  by  H.  From  Q  draw  perpendiculars  QX,  QY  to 
Q j^  PQ' and  PB.  The  displace- 
ment of  the  particle  Q  in 
time  dt  will  be  QY-  D,dt 
at  right  angles  to  the  plane. 
This  displacement,  however, 
can  also  be  obtained  by  com- 
pounding the  displacements 
produced  by  the  two  angular 
velocities  co,  co'.  That  produced  by  the  former  is  nil,  since  Q  is  on 
the  axis  of  rotation ;  that  produced  by  the  latter  is  QXco'dt.    Thus 

QY-  n  dt  =  QX-  ft)'  dt.  (117) 

We  have  QY-  PB  =  QX  -  PQ', 

each  being  equal  to  the  area  of  the  parallelogram,  and  on  combin- 
ing this  relation  with  (117),  we  find 


ft) 


n 


PB      PQ' 


ANGULAR  VELOCITY 


289 


Thus  if  ft)'  is  represented  Ly  I'(/,  then  H  moII,  on  tlie  same  scale, 
be  represented  by  PB. 

We  have  now  proved  the  following : 

The  resultant  of  two  angular  velocities  represented  hy  the  edges 
PQ,  PQ'  of  a  parallelograin  is  an  angular  velocity  represented  hy 
the  diagonal  PB  of  the  parallelogratn. 

Thus  angular  velocity  is  a  vector,  and  possesses  the  properties 
which  have  been  proved  to  be  true  of  all  vectors. 

232.  It  follows  that  an  angular  velocity  Vl  about  an  axis  of 
rotation  of  which  the  direction  cosines  are  I,  m,  n  may  be  replaced 
by  three  angular  velocities  oo-^,w^,  co^  about  the  axes  of  coordinates, 

&)j  = /n,  ft),  =  mfl,  ojg  =  Jili.  (118) 

Squaring  and  adding,  we  find  that 

Or  =  oj;  -f  &),-  +  col  (119) 

We  now  see  that  the  motion  of  a  rigid  body  is  given  when  we  know 
(ff)  u,  V,  %v,  the  components  of  velocity  of  the  point  P; 
(&)   ctjj,  fOj,  Wg,  the  components  of  angular  velocity. 


Kinetic  Energy  of  Eotation 

233.  Suppose  that  at  any  instant  a  rigid  body  is  rotating  about 
an  axis  of  rotation  PQ  with 
angular  velocity  H. 

Let  L  be  any  particle  of 
the  body,  its  mass  being  m, 
and  let  LN,  the  perpendicu- 
lar distance  from  L  to  PQ, 
be  denoted  by  p.  Then  the 
velocity  of  the  particle  L  is 
pQ,,  and  its  kinetic  energy  pj^  j^q 

is  ^mp'Dr. 

On  summation,  the  kinetic  energy  of  the  whole  bod}-  is  seen  to  be 


290  MOTION  OF   RIGID   BODIES 

The  quantity  '^injr  is  called  the  moment  of  iitertia  about  the 
axis  FQ. 

If  we  introduce  a  quantity  k,  defined  by 

2m]f 

^m 

so  that  A'^  is  the  mean  value  of  jl*^  averaged  over  all  the  particles  of 
the  body,  then  k  is  called  the  radius  of  gyration  about  the  axis  PQ. 
The  kinetic  energy  can  now  lie  written  in  the  form 

so  that  the  energy  is  the  same  as  if  the  whole  mass  were  concen- 
trated in  a  single  particle  at  a  distance  h  from  tlje  axis  of  rotation. 

Kinetic  Energy  of  a  Rigid  Body 

234.  The  point  P  is  at  our  disposal :  let  us  suppose  it  to  be  the 
center  of  gravity  of  the  body.  Then  the  most  general  motion  may 
be  compounded  of  a  motion  of  translation,  this  being  identical  with 
that  of  the  center  of  gravity,  and  a  motion  of  rotation  about  an 
axis  through  the  center  of  gravity. 

Let  V  be  the  velocity  of  the  center  of  gravity,  let  ft  be  tlie 
angular  velocity,  and  let  k  be  the  radius  of  gyration  about  the 
axis  of  rotation  through  the  center  of  gravity.  Let  M  be  the  total 
mass,  ^m,  of  the  body. 

By  the  theorem  of  §  186,  the  total  kinetic  energy  of  the  body  is 
the  sum  of  two  parts : 

{a)  the  kinetic  energy  of  a  single  particle  of  mass  M  moving 
with  the  center  of  gravity  of  the  body ; 

(b)  the  kinetic  energy  of  motion  relative  to  the  center  of  gravity. 

The  value  of  part  {a)  is  \  MV^ ;  that  of  part  {h)  is  \  Mk'n\ 
Hence  we  have  for  the  total  kinetic  energy 

^J/(r'+Fn2).  (120) 

This  expression  is  of  extreme  importance  in  itself,  but  is  also  of 
interest  because  it  enables  us  to  prove  the  following  theorem. 


KINETIC  ENEKGY  OF  A  RIGID  BODY 


291 


235.  Theorem.  Let  k  he  the  radius  of  gyration  ahout  any  axis 
through  the  center  of  gravity,  and  let  k'  he  the  radius  of  gyration 
ahout  a  j^arallel  axis  distant  a  from  the  former,  then 

k'^^l'+a\ 

Let  PQ  he  any  axis  through  the  center  of  gravity  G,  and  let 
P'  Q'  be  any  parallel  axis  distant  a  from  the  former.    Suppose  that 
the  rigid  body  has  a  motion  of  rotation 
about  P'Q',  the  angular  velocity  being  fl. 

Then  the  velocity  of  G  is  aVl,  and  the 
motion  may  be  regarded  as  compounded 
of  a  motion  of  translation  of  velocity  «n 
together  with  a  rotation  11  about  the 
axis  PQ.  By  formula  (120),  the  kinetic 
energy  is 

\M{am'+  JrD.% 

It  is  also  ^Mk"^D>^,  where  k'  is  the  radius  of  gyration  aljout  P'Q'. 
Hence  we  have 

^Mk'-n-=  ^M(a-n~+  z--n-), 

and  the  result  follows  on  di\'iding  through  h\  i  il/fi^ 

236.  Alternative  proof.   This  theorem  may  also  be  proved  geometrically. 

Let  L  be  any  particle  of  the  body,  and  let  the  plane  of  fig.  1-12  be 
supposed  to  be  the  plane  through  L  at  right  angles  to  the  two  axes  of 

rotation,  these  axes  cutting  the  plane  in 
the  points  ^4,  A'  respectively.  Let  LA  =  p, 
LA'  =  //.  and  let  LX  be  drawn  perpendicu- 
lar to  ^1.4'.     Then  Mk'^  =  ^;«y>-,  and  also 

3Ik"-  =  "^mp"' 

=  S"'  (/'"  +  --I'J"-  -  '^p  ■  .4.4'  cos  6) 
=  Mt^  +  .1/  •  .4 .4 '2  -  lM  a  '  {%m  ■  A  X). 

Now  AN  is  the  projection  of  the  line  from  L  to  the  center  of  gravity, 
upon  the  line  ^4,4'.    Hence  %m  ■  AX  =  0,  and  we  have 

il/r2  =  Ml-^  +  M-  ,4 .4 '2, 
giving  the  result  to  be  proved,  after  di\-ision  by  M. 


Fig.  142 


292  MOTION   OF  EIGID  BODIES 

237.  From  the  theorem  just  proved,  it  follows  that  the  radius 
of  gyration  about  any  axis  can  be  found  as  soon  as  we  know  that 
about  a  parallel  axis  through  the  center  of  gravity,  and  vice  versa. 
We  now  give  some  examples  of  the  calculation  of  radii  of  gyration. 


Calculation  of  Eadii  of  Gyration 

238.  Uniform  thin  rod.    Let  the  rod  AB  be  of  length  2  a,  and 

let  h  be  its  radius  of  gyration  about  an  axis  through  A  perpendic- 

„  „    ular  to  its  length.    Let  r  be  its  mass  per  unit 

A      X     P B  '^  ^ 

' '  1     length,  and  let  ir  be  a  coordinate  which  meas- 

FiG.  143  ^ 

ures  distances  from  A.  The  element  which 
extends  from  x  to  x  +  dx  i?,  of  mass  t  dx,  and  its  perpendicular 
distance  from  the  axis  of  rotation  is  x.    Hence 


3    '"     —  4  «2 


X2a 
(TfZ^)ar^ 

X'''  Crdx 

2a 
so  that  the  radius  of  gyration  is  — -  • 

^  V3 

About  the  center  of  gravity,  which  is  distant  a  from  A,  the 
radius  of  gyration  is  given  by 

^  3 

so  that  the  radius  of  gyration  about  the  center  of  gra\dty  is  —^  • 

239.  Rectangular  lamina.  Let  us  suppose  the  lamma  to  be  of 
edges  2  a,2h,  and  let  us  hud  its  radius  of  gyration  about  an  axis 
through  its  center  at  right  angles  to  its  plane.  Let  us  take  axes  as 
in  hg.  144,  and  let  o-  denote  the  mass  per  unit  area.    Then 


2^771/      J J(^  dxdy)  {^  +  if) 

ti,     ^^    — ~~ = 

Vi??,  4ab-  a- 


CALCULATION  OF  RADII  OF  GYRATION 


293 


The  integration  is  over  the  lamina,  and  therefore  between  the  limits 


a   to  X  =—  a   and   y  =h  to 
—  h.    On  integrating,  we  find 


k'  = 


3 


Fig.  144 


On  taking  h  =  0,  the  lamina 
becomes  a  thin  rod,  and  the  re- 
sult agrees  with  that  obtamed  in  the  last  section. 

240.  Homogeneous  solid  ellipsoid.  Let  the  semi-axes  of  the 
ellipsoid  be  a,  h,  c,  and  let  us  find  the  radius  of  gyration  about  the 
major  axis.  Taking  the  principal  axes  of  the  ellipsoid  as  axes  of 
coordinates,  and  denoting  the  density  of  the  ellipsoid  by  p,  we  have 


^mjy^ 


m 


j       {pdxdydz){y''+z-) 
, 

I  j  I  P dxdydz 


where  the  mtegration  is  over  the  whole  volume  of  the  ellipsoid. 
On  performing  the  integrations,  we  obtain 


EXAMPLES 

1.  Find  the  radius  of  gyration  of  a  rod  12  inche.s  long  about  a  point  distant 
4  inches  from  one  end. 

2.  Find  the  radius  of  gyration  of  a  circular  disk, 

(a)  about  an  axis  through  its  center  perpendicular  to  its  plane  ; 
(6)  about  a  diameter. 

3.  Show  that  the  radius  of  gyration  of  a  sphere,  radius  a,  about  any  diameter 
is  I  a''^,  and  about  any  tangent  line  is  l  a^. 

4.  What  is  the  radius  of  gyration  of  a  cube  about  an  edge  ? 

5.  What  is  the  radius  of  gyration  of  a  square  lamina  about  a  diagonal  ? 

6.  Find  the  radius  of  gyration  of  a  solid  circular  cylinder, 
(a)  about  an  axLs  ; 

(6)  about  a  generator ; 

(c)  about  a  diameter  of  one  of  its  ends. 

7.  Prove  that  the  radius  of  gyration  of  a  solid  conical  spindle  about  its  axis 
is  VI^  a,  where  a  is  the  I'adius  of  its  base. 


294  MOTION  OF  RIGID  BODIES 

Eouth's  Eule 

241.  The  following  convenient  rule,  given  by  Dr.  Eouth,  {Rigid 
Dynamics,  §  8),  provides  an  easy  way  of  remembering  the  values 
of  several  radii  of  gyration.  The  rule  applies  to  linear,  plane,  and 
solid  bodies  which  are 

(ft)  rectangular  (rod,  lamina,  or  parallelepiped) ; 

(b)  elliptical  or  circular  (disk  or  lamina) ; 

(c)  ellipsoidal,  spheroidal,  or  spherical  (solid) ; 

and  states  that  the  radius  of  gyration  about  an  axis  of  symmetry 
through  the  center  of  gravity  is  given  by 

,2  _  sum  of  squares  of  perpendicular  semi-axes 

A."      J 

3,  4,  or  5 
where  the  denominator  is  3,  4,  or  5  according  as  the  body  comes 
under  headings  («),  (b),  or  (c)  of  the  above  classitication. 

ILLUSTRATIVE  EXAMPLE 

A  coin  rolls  down  an  inclined  plane.    Find  its  velocity  after  any  distance  and 

also  its  acceleration. 

Let  the  coin  be  treated  as  a  uniform  circular  disk,  and  let  a  be  its  radius. 

When  its  velocity  down  the  plane  is  V,  its  angular  velocity  will  be  V/a.    The 

axis  of  rotation  is  perpendicular  to  the  plane  of 

the  coin.    Its  semi-axes  of  symmetry,  regarding  it 

as  a  lamina,  will  be  a,  a.    The  radius  of  gyration 

about  the  axis  of  rotation  through  its  center  is, 

by  Routh's  rule, 

a-  +  a- 

k^  = =  1  a2 

4 

so  that  the  kinetic  energy  is 


Fig.  145 


i  3/  r  F2  +  i  a2  (1  y  1  =  I MV^ 


After  rolling  a  distance  s  down  the  plane,  the  center  of  gravity  of  the  coin 
has  fallen  a  distance  s  sin  a,  so  that  from  the  conservation  of  energy 

Mgs  sin  a  =  1 3/F-, 
and  therefore  the  velocity  is  given  by 

F2  —  _4  gg  sin  a. 
Comparing  with  the  formula  (48),  V^  =  2/s,  for  motion  under  uniform  accel- 
eration, we  see  that  the  disk  rolls  down  the  plane  with  a  uniform  acceleration 
I  g  sin  a. 


MOMENT  OF  MOMENTUM  295 


EXAMPLES 

1.  Show  that  the  acceleration  of  a  hoop  rolling  down  a  hill  of  inclination  a 
is  J  g  sin  a. 

2.  Find  the  acceleration  of  a  pair  of  locomotive  wheels  running  down  a  gra- 
dient of  1  in  50,  each  wheel  consisting  of  a  rim  and  of  spokes  of  uniform  thick- 
ness, the  weight  of  the  rim  being  twice  that  of  the  spokes  and  the  weight  of  the 
axle  being  half  of  that  of  a  wheel.    (Neglect  the  thickness  of  the  axle.) 

3.  Two  bicyclists,  riding  exactly  similar  machines,  coast  down  a  hill,  start- 
ing with  equal  velocities  at  the  top.  Neglecting  the  forces  of  friction  and  the 
resistance  of  the  air,  show  that  the  heavier  rider  will  reach  the  bottom  first. 

4.  The  pulley  of  an  Atwood's  machine  is  a  uniform  disk  of  mass  M.  When 
masses  mi ,  rrii  ax'e  attached  to  the  ends  of  the  string,  show  that  the  acceleration 
of  nil  is  ^        ^ 


mi  +  m2  +  ^M^ 


5.  Two  spheres,  one  a  hollow  shell  and  the  other  a  homogeneous  solid,  roll 
down  hill  together,  starting  simultaneously  from  rest  at  the  top.  Show  that 
their  times  over  any  part  of  the  path  are  in  the  ratio  5  :  V 21. 

6.  If  the  masses  of  the  wheels  of  a  carriage  are  supposed  to  be  all  collected 
at  the  rim,  show  that  the  energy  of  the  carriage  when  moving  with  velocity  V 
is  h  MV^,  where  M  is  the  weight  of  the  complete  carriage  plus  the  weights  of 
the  wheels. 

7.  A  straight  piece  of  uniform  wire  is  stood  vertically  on  end  and  allowed  to 
fall  over.    With  what  velocity  does  it  strike  the  ground  ? 

8.  A  homogeneous  solid  cigar-shaped  spheroid,  semi-axes  a  and  b,  is  stood 
on  its  point  on  a  horizontal  plane  and  is  allowed  to  roll  over.  Find  its  angular 
velocity  when  the  end  of  its  minor  axis  is  in  contact  with  the  plane,  and  find 
the  pressure  on  the  plane  at  this  instant. 


Moment  of  Momentum 

242.  Let  X,  y,  z  be  the  coordinates  of  any  particle  of  mass  m. 
Let  the  components  of  the  total  resultant  force  acting  on  the  par- 
ticle be  X,  Y ,  Z.    Then  the  equations  of  motion  are 


290 


MOTION  OF  RIGID  BODIES 


The  moment  about  the  axis  of  x  of  the  force  actmg  on  the 
particle  is  yZ—  zY,  and  from  the  foregoing  equations  we  have 


yZ—  zY  =  only 


cW 


cW 


(121) 


The  velocity  of  the  particle  has  components  -r '  -r^  >  —  >  so  that 
^  ^  ^  dt    dt    dt 

the  moment  of  this  velocity  about  the  axis  of  z,  as  defined  in 

S  "^     '  dz         dy 

It' 


^  dt 


The  momentum  of  the  particle  is  m  times  its  velocity,  so  that 
the  moment  of  momentum  about  the  axis  of  z  is  m  times  the 
moment  of  the  velocity,  and  therefore 


dz 


dy 


^^  dt      ^  dt 


On  differentiating,  we  have 


dt 


m 


dz         dy 


=  m 


dy  dz  d^z 

—  — \-  y  — 

dt  dt  dt 


d'^z         d'^y 


dz  dy         d?y 
dt  dt  df 


df 


(122) 


de 
=  yZ-zY, 

by  equation  (121).    Thus  we  have  proved  that 

The  rate  of  change  of  the  moment  of  momentum^  of  a  particle 
about  any  axis  is  equal  to  the  moment,  about  the  same  axis,  of  the 
forces  acting  on  the  ^particle. 

243.  Equation  (122)  is  true  for  each  particle  of  any  system  of 
bodies.    Let  us  sum  the  equation  over  all  particles,  then  we  obtain 


d_ 

dt 


m 


dz 


<^y 


^  dt      ^  dt 


■■X{}JZ 


zY 


(123) 


The  right-hand  side  of  this  equation  is  the  sum  of  the  moments 
of  the  external  forces  acting  on  the  body  or  system  of  bodies, 


MOMENT  OF  MOMENTUM  297 

for  the  internal  forces  occur  in  equal  and  opposite  pairs  wliich 
contribute  nothing. 

The  term  V??i  l  y~  —  z-^)>  which  is  the  sum  oi  the  moments 
^     y  dt         cltj 

of  momentum  of  the  separate  particles,  is  called  the  moment  of 
momientum  of  the  system. 

Thus  equation  (123)  expresses  that 

The  rate  of  ehange  of  the  moment  of  momentum  of  any  system 
about  any  axis  is  equal  to  the  sum  of  the  moments  of  the  external 
forces  about  this  axis. 

244.  Several  important  consequences  of  this  theorem  follow 
at  once. 

\.  If  a  system  of  bodies  is  acted  on  by  no  external  forces,  the 
moment  of  momentum  about  every  axis  remains  constant. 

This  expresses  the  prmciple  known  as  the  conservation  of  angu- 
lar momentum. 

The  sun  affords  an  instance  of  a  body  which  may  practically  be  sup- 
posed to  be  acted  on  by  no  external  forces.  It  is  generally  supposed  that 
the  sun  is  gradually  shrinking  in  size ;  if  this  is  so,  we  see  that  its  velocity 
of  rotation  about  its  axis  must  continually  increase,  in  order  that  its 
moment  of  momentum  may  remain  constant. 

II.  If  all  the  forces  acting  on  a  system  are  either  parallel  to  a 
given  line,  or  else  intersect  this  line,  then  the  moment  of  momentum 
of  the  system  about  this  line  7)iust  remain  constant. 

A  peg  top  is  acted  on  only  by  the  reaction  at  the  peg  and  gravity.  Tlie 
moment  of  the  latter  about  a  vertical  line  through  the  peg  vanishes,  and 
the  moment  of  the  former  may  be  sujiposed  to  vanish  to  a  close  approxi- 
mation. Hence  the  moment  of  momentum  about  a  vertical  through  the 
peg  will  remain  constant,  to  a  close  approximation. 

III.  If  a  rigid  body  is  free  to  rotate  about  a  fixed  axis,  and  if 
ft)  is  its  angular  velocity  at  any  instant,  then 

dt 
where  MP  is  the  moment  of  inertia  about  the  fixed  axis,  and  L  is 
the  sum  of  the  moments  about  this  axis,  of  all  the  external  forces. 


298  MOTION   OF  EIGID   BODIES 

To  see  the  truth  of  this,  it  is  oiil}'  necessary  to  notice  that  a  par- 
ticle of  mass  m  at  distance  p  from  the  axis  has  momentum  mpw, 
so  that  the  moment  of  momentum  of  the  whole  system  will  be 

and  smce  M  and  k"^  do  not  vary  with  the  time,  the  rate  of  change 
of  angular  momentum  will  be  3IJi^  —  • 


Oscillation  of  a  Pendulum 

245.  An  important  application  of  the  last  theorem  enables  us 
to  find  the  time  of  oscillation  of  a  pendulum  of  any  description. 

Let  0  be  the  pivot  about  which  the  pendulum  turns,  let  G  be 
its  center  of  gravity,  \Qi  OG  =  h,  and  let  the  hue  OG 
make  an  angle  6  with  the  vertical  at  any  instant,  so 

that  ft)  =  —  is  the  angular  velocity  of  the  pendulum 

about  its  axis. 

Let  M  be  the  mass, -and  k  the  radius  of  gyration 
about  its  axis,  of  the  whole  pendulum.  Then  the 
equation  of  motion  is 


T^7Q    Cl^  T 

Ml?  -—  =  L, 

dt 


dd 


Fig.  146      [^   which   &)  =  —  •    The  value  of  L  is  equal  to  the 

dt 

moment  of  the  weight  about  the  axis  through  0,  and  is  therefore 

Mgh  sin  9. 

Thus  the  equation  becomes 

Mk^^  =  M(ih  sine, 

Pd'e       .  a 

or,  --  — — -  =  a  sm  u. 

h  dt'      -^ 


MOMENT   OF  MOMENTUM  299 

The  equatiou  of  luution  for  a  simple  pendulum  of  length  I  is 

Z-=^sm^, 

so  that  we  see  on  comparison  that  the  motion  is  the  same  as  that 
of  a  simple  pendulum  of  length  I  =  k'^/h. 

For  instance,  the  complete  period  of  small  osciLLations  is 


27r 


(jh 


ILLUSTRATIVE  EXAMPLE 

A  ring  (e.g.  a  dinner  napkin  ring)  stands  vertically  on  a  table,  and  a  gradually 
increasing  pressure  is  applied  by  a  finger  to  one  point  of  the  ring  in  such  a  way 
that  equilibrium  is  broken  by  the  point  of  contact  with  the  table  slipping  along  the 
table.    Find  the  subsequent  motion  of  the  ring. 

We  have  seen  in  example  2,  p."  109,  that  it  is 
pos.sible  to  apply  pressure  in  the  manner  described. 

Let  us  suppose  that  when  the  ring  leaves  the 
finger  it  is  observed  to  be  moving  with  a  velocity 
V  forward  and  a  rotation  ii  in  the  direction  oppo- 
site to  that  in  which  it  would  rotate  if  it  were 
rolling  without  sliding.  Let  v,  w  be  the  values  of 
the  velocity  and  rotation  at  any  instant,  measured 
in  the  same  directions  as  V  and  12. 

Let  a  be  the  radius  of  the  ring  and  m  its  mass, 

(a)  its  weight  mg  ; 

(6)  the  vertical  component  of  the  reaction  with  the  table,  which  is  ecjual  to 
mg  since  the  center  of  gravity  of  the  ring  has  no  vertical  acceleration  ; 

(c)  the  frictional  reaction  at  the  lowest  point  of  the  ring,  which  is  equal  to 
mg  n  so  long  as  sliding  takes  place. 


Fig.  147 
The  forces  acting  on  it  are 


By  the  theorem  of  §  180  we  have 


dv 
di 


-  rng/J.. 


(a) 


We  can  obtain  a  second  equation  from  the  theorem  of  §  243.  Let  us  take  as 
axis  the  axis  of  the  ring  at  instant  t.  The  moment  of  inertia  at  this  instant  is 
7na2.  To  obtain  the  moment  of  momentum  we  regard  the  whole  motion  as  com- 
pounded of  a  motion  of  translation  of  the  center  of  gi-avity  (velocity  v),  and  a 


GIFT  OF 


DR,  GEORGE  F.  McEWEN 
MC 


300  IroTIOK  OF  RIGID  BODIES 

motion  of  rotation  about  an  axis  through  the  center  of  gravity  (velocity  w). 
The  former  contributes  nothing  to  the  moment  of  momentum,  so  that  the  whole 
moment  of  momentum  is 

?«a-c<;. 

At  the  end  of  a  small  interval  dt  the  ring  will  have  moved  forward  a  distance 
vdt,  so  that  we  are  now  considering  the  moment  of  inertia  about  an  axis  which 
is  distant  v  dt  from  the  center  of  gravity  of  the  ring.  The  moment  of  inertia 
after  an  interval  dt  is,  accordingly,  by  §  235, 

m(a'^  +  {vdty-). 

We  may,  however,  neglect  the  small  quantity  of  the  second  order  (dt)^  and 
treat  the  moment  of  inertia  as  though  it  remained  constant  and  equal  to  ma"^. 

The  rate  of  increase  of  the  moment  of  momentum  is,  accordingly   rna^  —  • 

dt 
The  moment  of  the  external  forces,  measured  about  the  same  axis  and  in  the 
same  direction,  is 

-  mg  fxa, 

d<i3 
so  that  we  have  the  equation    1710,"^  —  =:—  mg na,  (b) 

or,  simplified,  a  —  =  —  /xg,  (c) 

while  equation  (a)  reduces  to  —  =  —  f^g.  (d) 

dt 

These  relations  give  the  rates  of  decrease  of  v  and  w  so  long  as  sliding  is 
taking  place.  Sliding  clearly  ceases  as  soon  as  we  have  v  +  wa  =  0,  for  v  +  wa 
is  the  forward  velocity  of  the  lowest  point  of  the  ring.  Fi'om  equations  (c)  and 
(d)  we  have 

—  (?)  +  wa)  =  -  2  fig, 
dt 

and  initially  the  value  of  v  +  wa  is  V+  Oa.  The  time  required  to  reduce  v  +  wa 
to  zero  is,  accordingly, 

V+  ila 

After  this  interval  sliding  ceases.  The  velocity  of  the  ring  at  this  instant  is 
given  by 

v=  V-fig-[~- ) 

=  l{V-na), 

so  that  the  motion  may  be  either  forwards  or  backwards  according  as  we  had 
initially  F>  or  <  fia.  After  sliding  has  once  ceased  there  is  no  force  tending 
to  start  it  afresh,  so  that  the  ring  simply  rolls  on  with  uniform  velocity  v.  If 
V>Ua,  it  rolls  farther  from  its  point  of  projection;  while  if  V<iia,  it  will 
return  to  the  point  of  projection. 


GENERAL  THEORY  OF  MOMENTS  OF  INERTIA     301 

EXAMPLES 

1.  The  line  of  hinges  of  a  door  makes  an  angle  a  with  the  vertical,  and  the 
door  swings  about  its  position  of  equilibrium.  Show  that  its  motion  is  the  same 
as  that  of  a  certain  simple  pendulum,  and  find  the  length  of  this  pendulum. 

2.  A  target  consists  of  a  square  plate  of  metal  of  edge  a  and  of  mass  3f, 
hinged  about  its  highest  edge,  which  is  horizontal.  When  at  rest  it  is  struck  by 
an  inelastic  bullet  of  small  mass  m  moving  with  velocity  v,  at  a  point  at  depth  h 
below  the  line  of  hinges.    Find  the  subsequent  motion  of  the  target. 

3.  A  homogeneous  sphere  is  projected  without  rotation  up  a  rough  inclined 
plane  of  inclination  a  and  coefficient  of  friction  n.  Show  that  the  time  during 
which  the  sphere  ascends  the  plane  is  the  same  as  if  the  plane  were  smooth,  and 
that  the  time  during  which  the  sphere  slides  stands  to  the  time  during  which  it 
rolls  in  the  ratio  7  /n  :  2  tan  a. 

4.  A  sphere  of  radius  a  is  held  at  rest  at  a  point  on  the  concave  surface  of 
a  spherical  bowl  of  radius  b.  It  is  suddenly  set  free  and  allowed  to  roll  down 
the  surface.  Show  that  the  line  joining  the  centers  of  the  tw^o  spheres  swings 
in  the  same  way  as  a  simple  pendulum  of  length  I  {h  —  a). 

5.  A  sphere  of  radius  a  is  held  at  rest  at  the  highest  point  of  the  rough  con- 
vex surface  of  a  sphere  of  radius  b.  It  is  then  set  free  and  allowed  to  roll  down 
this  sphere.  Show  that  the  spheres  will  separate  when  the  line  joining  their 
centers  makes  an  angle  cos-i|^  with  the  vertical.    Examine  the  case  of  6  =  0. 

6.  A  circular  hoop,  which  is  free  to  move  on  a  smooth  horizontal  plane,  has 
sliding  on  it  a  small  ring  of  1/nth  its  mass,  the  coefficient  of  friction  between 
the  two  being  /i.  Initially  the  hoop  is  at  rest,  and  the  ring  has  an  angular 
velocity  w  round  the  hoop.    Show  that  the  ring  comes  to  rest  relative  to  the 

hoop  after  a  time 

General  Theory  of  Moments  of  Inertia 

Coefficients  of  Inertia 

246.  Suppose  that  a  rigid  body  is  rotating  about  an  axis  of 
rotation  of  which  the  direction  cosines,  referred  to  any  three  fixed 
coordinate  axes,  are  /,  m,  n.  Let  us 
take  any  pomt  O  on  the  axis  of  rota- 
tion for  origin,  and  let  L  be  any  par- 
ticle of  mass  m^,  distant  p  from  the 
axis  of  rotation.  Let  the  coordinates 
of  L  be  X,  y,  z,  and  let  LN(=p)  be 
the  perpendicular  from  L  on  to  the 
axis  of  rotation.  •"  fiq.  143 


302  MOTION   OF  KIGID    I50DIES 

We  have       OL''  =  x"  +  if  +  z-, 

ON''={lx  +  vii/  +  nz)\ 
so  that  f=OL''-ON' 

=  x^  +  y^  -\-  z-  —  (Ix  +  my  +  nz)^ 

=  x^  {7n'  +  n-)  +  f  {iv"  +  /')  +  z'  (/■'  +  vir) 

—  2  mn  •  yz—2nl  •  zx  —  2  bn  ■  xy 
=  /-  if  +  z^)  +  ??i'  (2;''  +  x^)  +  w^  (^'^  +  y") 

—  2  7/1;;,  •  yz—lnl  •  zx  —  llm  -  xy. 

Hence  the  moment  of  inertia,  say  /,  is  given  by 

—  2  inn  "^m^yz  —  2  ?i/  '^ni-^zx  —  2  /'>?z-  V?/i;^a-y 
=  l^A+  m-B  +  ?r  C  -  2  mnD  -  2  7i/^  -  2  Zmi^,  (124) 

where  ^  =  ^^'^i  (i/^  +  s^),  etc., 

D  =  '^in^yz^  etc. 

The  quantities  A,  B,  C  are  seen  to  be  the  moments  of  inertia 
about  the  axes  of  x,  y,  z  respectively.  The  quantities  D,  E,  F  are 
called  pi'oducts  of  inertia. 

By  giving  different  values  to  /,  m,  n  in  equation  (124),  we  can 
find  the  moment  of  inertia  about  any  line  through  0,  as  soon  as 
we  know  the  values  of  the  six  coefficients  A,  B,  C,  D,  E,  F. 

Ellipsoid  of  Inertia 
247.  The  equation 

Ax''  +  By"  +Cz'—2  Dyz  —  2  Ezx  -  2  Fxy  =  K, 

where  K  is  any  constant,  being  of  the  second  degree,  represents  a 
conicoid.    If  r  is  the  radius  vector  of  direction  cosines  /,  m,  n,  we  have 

r-  {AV  +  Bm}  +  Cn''  —  2  Dmn  —  2  Enl  —  2  Flm)  =  K, 
or,  from  equation  (124),  ^^2__.  (125) 


Since  /  is  positive  for  all  values  of  /,  m,  n,  it  follows  that  r^  is 
positive  for  all  directions  of  the  radius  vector.  Thus  the  conicoid 
is  seen  to  be  an  ellipsoid. 

This  ellipsoid  is  called  the  ellijysoid  of  inertia  of  the  point  0. 

Equation  (125)  may  be  written 

and  now  expresses  that  the  moment  of  inertia  about  any  axis 
through  0  is  inversely  proportional  to  the  square  of  the  parallel 
radius  vector  of  the  ellipsoid  of  inertia. 


Principal  Axes  of  Inertia 

248.  This  physical  property  of  the  ellipsoid  shows  that  the 
ellipsoid  itself  remains  the  same,  no  matter  what  axes  of  coor- 
dmates  are  chosen.  The  ellipsoid  has  three  principal  axes,  wliich 
are  mutually  at  right  angles.  The  directions  of  these  axes  are  called 
the  principal  axes  of  mertia,  at  the  point  0. 

If  the  principal  axes  of  mertia  at  the  point  0  are  taken  as  axes 
of  coordinates,  then  the  coetiicients  of  yz,  zx,  xy  in  the  equation  of 
the  ellipsoid  must  disappear.    Thus  w^e  must  ha,ve 

D  =  E  =  F=Q. 

Taking  the  principal  axes  of  inertia  at  0  as  axes  of  coordmates, 
equation  (124)  assumes  the  form 

I=PA  +  m-B  +  n-C. 
The  kinetic  energy  of  a  rotation  of  angular  velocity  H  is 

1  /  li-  =  ^  {PA  +  vr  B  +  li-  C)  n- 

=  \  {A(o'l  +  Bo}.^  +  fW) ,  (126) 

where  (o^,  co.-,,  cOg  are  the  components  of  11  (see  §  232). 


304 


MOTION   OF  EIGID   BODIES 


General  Equations  of  Motion  of  a  Rigid  Body 

249.  Let  0  be  any  point  of  a  rigid  body,  and  let  Ox,  Oy,  Oz  be 
a  set  of  axes  moving  so  that  the  point  O  maintains  its  position  in 
the  rigid  body,  while  the  axes  remain  parallel  to  their  original 

position. 

Let  the  velocity  of  0  have  compo- 
nents u,  V,  w  along  these  axes.    The 
motion  of  the  rigid  body  relative  to 
.these  axes  will  be  a  motion  of  rotation 
about  some   axis  OP  which   passes 
through  0.    Let    us   regard   this  as 
compounded  of  rotations  to^,  w^,  w^ 
about  the  three  axes. 
Let  X,  y,  z  be  the  coordinates  of  any  point  of  the  rigid  body  rela- 
tive to  these  axes.    The  velocity  of  this  point  relative  to  the  frame 
supplied  by  the  axes  moving  with  0  has  components 


Fig.  149 


dx 
lit 


dy 
dt 


dz 
di. 


wliile  the  velocity  of  this  frame  in  space  has  components 

u,  V,  w. 

Thus  the  whole  velocity  of  the  point  x,  y,  z  has  components 

dz 


dx  dy 

dt  dt 


%v  -f- 


dt 


At  any  instant  let  L,  M,  N  denote  the  sums  of  the  moments  of 
the  external  forces  about  the  axes  of  x,  y,  z  respectively,  so  that,  as 
in  §  243, 

L  =  ^{yZ-zY),Qtc. 

The  moment  of  momentum  of  a  particle  of  mass  m,  at  x,  y,  z, 
about  the  axis  of  x  is 


m 


dz 


^"-1 


GENERAL   EQUATIONS  OF  MOTION" 


305 


Hence,  by  the  theorem  of  §  243, 


dt 


dz\  (        dii 


=z, 


(127) 


and  there  are  similar  equations  for  the  otlier  axes. 

250.  Eelative  to  the  moving  axes  of  coordinates  the  particle  m 
has  coordinates  x,  y,  z,  so  that  a  rotation  w^  about  Ox  gives  the 
particle  a  velocity  of  components 

0,  -  oi^z,         (ojj. 

Similarly  the  rotations  to^,  w^  give  velocities  respectively  of 
components  n 

and  —  cD.y,  wjc^  0. 

Compounding  these  velocities,  we  obtain  as  the  components  of 
the  resultant  velocity,  relative  to  the  axes, 

dx 
'di 
dy 


^„^  -  ^-^y^ 


Thus 


dz 


II z 

^  dt  dt 


-  =  co^x-a>^, 


dz  _ 

di~ 

dy  _ 


<»x{if  +  ^)  —  <^y  yy  —  (o^xz. 


and  on  differentiation  of  this  equation  with  respect  to  t,  we  obtain 
as  the  value  of  part  of  the  left-hand  member  of  equation  (127) 


d_ 
dt 


% 


m[y 


dz 
~dt 


dy 
di 


■■%'"<f  +  ^)'^-%'«-^/^-X 


VI  xz 


dco. 


dt       ^       ''    dt       ^  dt 

-^m  ?/z{co;  -  (o;)  +^"i(/  -  5r)tw^a), 

— V/^i  zx  (OyCO_.  +^m  :>y  w,w_^ 


f/co,.  d(o  dxo^ 

A  — i  —  F  — '-  —  E  — - 

dt  dt  dt 


—  D{Wy  —  w':)  —  {B  —  C)  CO  (1)^  —  E (o^j03^.  +  F (jo.(o^ 


306  MOTION  OF  EIGID  BODIES 

251.  Let  X,  y,  z  be  the  coordinates  of  the  center  of  gravity  of 
the  rigid  body,  and  let  M  be  its  total  mass.    Then 

V>/(,:f  =  J/,Z',  etc. 

As  the  value  of  the  remaining  part  of  the  left-hand  member  of 
equation  (127)  we  now  have 

=  M —  (yw  —  zv). 
Thus  equation  (127)  now  assumes  the  form 

M—  (yw  —  zv)+A  — ^—F — 'J  —E  — '  —Dlu)-  —  co:) 
dr^  '  dt  dt  dt  \  y        -I 

-{B-  C)  w^(D^  -  Ew,w^  +  Fco^co^  =  L.         (128) 

If  ^A\  ^y,  2j^  denote  the  total  components  along  the  axes, 
we  have,  by  §  180,  the  further  equations 

^4(»+f)-5;-v-  (129) 


dt  \         dt ^ 

Equations  (128)  and  (129)  and  the  two  other  pairs  of  equations 
corresponding  to  the  two  other  axes  are  the  equations  of  motion 
for  a  rigid  body  moving  under  any  forces. 

Euler's  Equations 

252.  Let  us  now  suppose  that  we  have  a  second  set  of  axes, 
which  we  shall  denote  by  1,  2,  3.  Let  these  axes  move  so  as 
always  to  retain  the  same  position  in  the  rigid  body,  the  point  O 
(which  we  have  already  supposed  always  to  retain  the  same  posi- 
tion in  the  rigid  body)  being  the  origin.  Let  the  axes  1,  2,  3  coin- 
cide with  the  axes  .r,  ?/,  z  at  the  instant  under  consideration.  Then 
the  values  of  the  coefficients  of  inertia  referred  to  axes  1,  2,  3  are 
the  same  as  those  referred  to  axes  x,  y,  z,  namely  A,  B,  C,  D,  E,  F. 


EULER'S  EQUATIONS  307 

Moreover,  all  velocities  referred  to  axes  1,  2,  3  have  the  same  val- 
ues as  they  would  have  if  referred  to  axes  x,  y,  z.  Let  us  denote 
the  rotations  about  the  axes  1,  2,  3  by  oo^,  w„,  w^,  then  at  the 
mstant  under  consideration  we  shall  have 


This  is  not  necessarily  true  at  any  instant  except  the  instant  at 
which  the  axes  coincide,  so  that  it  is  not  permissible  to  differen- 
tiate these  equations  with  respect  to  the  time  and  deduce  that 

^^=^^etc. 
at        at 

Nevertheless,  it  can  be  shown  that  tliis  last  result  is  true  at  the 
instant  under  consideration.  Let  OQ  denote  any  Hne  through  O, 
let  cos  a,  cos  ^,  cos  7  be  its  direction  cosines  relative  to  axes  1,  2,  3, 
and  let  12^  be  the  component  of  angular  velocity  about  OQ.  If  the 
resultant  angular  velocity  is  one  of  amount  H  about  an  axis  OP 
of  which  the  direction  cosines  referred  to  axes  1,  2,  3  are  /,  m,  n, 
then  we  have 

fi,^=  n  cos  POQ 

=  il(l  cos  a  +  w  cos  a  +  n  cos  /3) 
=  (Bj  cos  a  -f-  &)„  cos  /3  -t-  0)3  cos  7. 

Whatever  line  OQ  may  be,  this  equation  is  always  true ;  hence  we 
may  legitimately  differentiate  it  with  respect  to  the  time,  and  so 
obtain 

dfl^      d(o,  d(o„        ^      d(o^ 

— 2  =  — i  cos  a  -\ cos  p  -I cos  7 

dt         dt  dt  dt  ' 

da  .     „d^  .        ^/7  ,.^^^ 

—  cu^smn;--  —  a)^  sm/3-— —  0)3  sm  7-— ■       (loO) 
dt  at  dt 

Now  let  the  luie  OQ  he  supposed  to  coincide  with  Ox,  so  that 

n^  =  (D^.    At  the  instant  under  consideration,  /8  =  7  =  —  -  a'  =  0. 

dB  ~ 

Moreover,  — -  is  the  rate  at  which  the  angle  between  Ox  anil  axis 

dy  d(X 

1  increases,  and  clearly  this  is  w^.    Similarly,  — ^  =  —  w.,  antl  ^  =  0. 

dt  '         dt 


«2-«3  +  «^3<y2 


308  MOTION   OF   RIGID   BODIES 

Making  all  these  substitutions,  we  find  that  at  the  moment  under 
consideration,  at  which  the  two  sets  of  axes  coincide,  equation  (130) 
assumes  the  form 

dt         dt 
dco^ 
"'dt  ' 

Thus  at  the  instant  at  which  the  two  sets  of  axes  coincide,  we 

have  the  relations 

a)j  =  (Wj,  etc., 

and  also  -^  =  ^7  • 

dt         dt 

Let  us  introduce  the  further  simplification  of  supposing  that  the 
origin  is  either  a  fixed  point  or  the  center  of  gravity  of  the  body- 
In  the  former  case  we  have 

tc  =  V  =  w  =  0,  always  ; 
in  the  latter  case 

X  =  1J  =:z=  0,  always. 

Let  us  further  suppose  that  the  axes  are  chosen  to  be  the  prin- 
cipal axes  of  inertia  through  the  origin,  so  that 

I)  =  E  =  F=0. 

Introducing  all  these  simplifications  into  equation  (128)  and  the 
two  similar  equations,  we  find  that  these  assume  the  form 

a'^-(B-C)co.^o),=L,  (131) 

dt 

b'!^--(C-A)  a>,co,  =  31,  (132) 

dt 

C"^-  {A-B)co^co„  =N.  (133) 

dt 

These  equations  are  known  as  Euler's  equations. 


ROTATION  OF  A  PLANET  309 

Rotation  of  a  Planet 

253.  As  a  first  example  of  the  use  of  these  equations,  let  us 
examine  the  motion  of  a  rigid  body,  symmetrical  about  an  axis, 
acted  on  by  forces  all  of  wiiich  pass  through  the  center  of  gra^•ity. 
These  conditions  approximately  represent  those  which  obtain  when 
a  planet  moves  in  its  orbit,  or  a  star  in  space. 

Let  us  take  the  center  of  gravity  as  origin  and  the  axis  of  sym- 
metry as  axis  1.  Let  the  moments  of  inertia  be  A,  B,  B.  Then  the 
equations  of  motion  are 

a'^  =  0.  (134) 

B^  =  {B-A)cc,<o,,  (135) 

^^  =  -^^-^^)''^''-  (136) 

The  first  equation  gives  at  once  that  co^  is  constant,  say  equal 
to  II.    If  we  write 

B 

equations  (135)  and  (136)  become 

^'  =  ^™"  (137) 

Ihus  — -=  =  k  — -  =  —  Pa)2, 

df  dt 

of  which  the  solution  is     &),,  =  E  cos  {kt  -^  e) ; 
and  equation  (137)  now  leads  at  once  to 

(Wg  =  —  J£'sin(/.'/  +  e). 

Thus  the  components  of  angular  velocity  at  the  instant  t  are 
n,  E  cos  {Id  +  e),  E  sin  {Jd  4-  e), 


310  MOTION   OF   lUGIl)   BODIES 

and  we  see  that  the  axis  uf  rotation  describes  a  cone  in  the  solid, 

withpenod  —  or— -^-^- 

If  ^  is  very  nearly  equal  to  A,  the  period  may  be  very  great, 
and  the  motion  consequently  very  slow.  This  happens  in  the  case 
of  the  earth :  the  motion  of  the  axis  of  rotation  gives  rise  to  the 
phenomenon  known  as  the  variation  of  latitude,  of  which    the 

2  TT 

period  is  about  428  days.    Since  a  period  — -  represents  roughly 

B—A 

one  day,  we  conclude  that  for  the  earth  — — —  is  of  the  order  of  ^|g. 

The  true  value  of  this  quantity  is  .00328,  the  discrepancy  resulting  from 
the  imperfect  rigidity  of  the  earth. 

Motion  of  a  Top 

254.  As  a  second  example  of  the  methods  of  this  chapter,  let  us 
consider  the  motion  of  a  spinning  top.  This  we  shall  suppose  to 
be  a  solid  of  revolution  spinning  on  a  peg  of  which  the  end  will  be 

treated  as  a  point,  the  contact  between 
the  peg  and  the  surface  on  which  it 
rests  being  assumed  rough  enough  to 
prevent  slipping.  The  point  of  contact 
is  now  a  fixed  point  0.  Let  us  take 
axes  Ox,  Oy,  Oz  fixed  in  space,  the  axis 
of  z  being  vertical,  and  also  axes  I,  2,  3 
fixed  in  the  body,  and  coinciding  with 

the  principal  axes  of  inertia  through  0. 

Fig.  150  -r    .         ■      i    i       ^i  •        £  ^  f 

Let  axis  1  be  the  axis  of  symmetry  oi 

the  top,  and  let  the  moments  of  inertia  about  axes  1,  2,  3  be  ^,  ^,  i?. 

The  first  of  Euler's  equations  becomes 

^^  =  0' 

at 

since  B  =  C  and  i  =  0.    Thus  cwj  is  a  constant,  say  fl. 

Let  the  axis  of  the  top  cut  a  unit  sphere  about  O  at  a  point  whose 
polar  coordinates  are  1,0,  (f),  the  axis  of  Oz  being  taken  for  pole,  so 
tliat  6  is  the  angle  between  the  vertical  and  the  axis  of  the  top. 


MOTION  OF  A  TOP  311 

The  kinetic  energy  of  the  top  is,  by  §  248, 

while  the  potential  energy  is  3I(/h  cos  6,  where  h  is  the  distance 
of  the  center  of  gravity  of  the  top  from  O.  Thus  the  equation  of 
energy  is 

An^  +  B  ((4  +  (ol)  +  2  Mgh  cos  6  =  B,  (139) 

where  E  is  a  constant.  Tliis  may  be  put  mto  a  different  form. 
For  0)1  +  (ol  is  the  square  of  the  angular  velocity  of  the  axis  of 
the  top :  it  is  therefore  the  square  of  the  actual  velocity  of  the 
point  1,  ^,  ^  on  the  unit  sphere,  and  hence  we  have 

The  equation  of  energy  now  assumes  tlie  form 


An^  +  B 


fJ-"-(fJ 


+  2  3fghcosd  =  E.     (140) 


We  can  obtain  a  tliird  equation  from  the  fact  that  the  angular 
momentum  about  Oz,  the  vertical,  is  constant.  The  angular  momen- 
tum may  be  regarded  as  compounded  of 

(a)  the  momentum  due  to  the  rotation  H  about  axis  3 ; 

(b)  the  momentum  due  to  tlie  motion  of  the  axis  of  the  top. 

The  rotation  fl  about  axis  3  may  be  further  decomposed  into 
rotations  fl  cos  6,  D.  sm  6  about  the  horizontal  and  vertical,  givmg 
moments  of  momenta  AD,  cos^,  AD,  sin^  about  the  horizontal  and 
vertical.  Thus  the  moment  of  momentum  contributed  by  part  (a) 
is  AD  cos  d. 

The  motion  of  the  axis  of  the  top  may  be  resolved  into  a  rota- 
tion of  angular  velocity  sin  6  -j-  about  an  axis  making  an  angle 

—  —  ^  with  the  vertical,  and  one  of  angular  velocity  —  about  a 
^  etc 


312 


MOTION   OF  RIGID   BODIES 


dcl> 


horizontal  axis.    The  former  may  be  replaced  by  sin-  6  --   about 

7    1  (U 

the  vertical,  and  sin  6  cos  6  -—■  about  a  horizontal  axis.    Thus  the 

at 

moment  of  momentum  about  the  vertical  contributed  by  part  (//) 
of  the  motion  is 


^sin-^ 


dt  ' 


and  since  the  moment  of  momentum  about  the  vertical  has  a  con- 
stant value,  say  G,  we  have 


^acos6'  +  i?sin--^6'^  =  (?. 
dt 

d(f> 


(141) 


If  we  eliminate  — ^  from  tliis  equation  and  equation  (140),  we 
obtain 


^sin=^6^ 


An^  +  B{^^\+2Mghcos0-E 


+  (G -An  cos  6)- =  0,  (142) 
a  differential  equation  giving  the  variations  in  the  value  of  6,  and 
therefore  allowing  us  to  trace  the  changes  in  the  inclination  of  the 
axis  of  the  top  to  the  vertical. 

The  maxima  and  minima  of  6  are  given  by  putting  -7-  =  0,  f^iid. 
are  therefore  the  roots  of 

^  (1  -  cos-  d)  [An^  +  2  3I(/h  cos  O-JiJ] 

-i-{G-Ancose)'  =  0.     (143) 

Let  us  call  the  left  hand  of  this  equation /(cos  0).  Since/  is  a 
function  of  degree  three,  there  will  be  three  roots  for  cos  6.  Let 
us  suppose  that  the  top  is  started  at  an  angle  0=6^,  and  with  the 

value  of  —  equal  to  ( -7-  )•    Then,  from  equation  (142), 


^sm=^^„ 


An^  +  2?  (  ^  Y  +  2  3Igh  cos  6>,  -  E 


+  (G  -  An  cos  e,y  =  0, 

so  that        / (cos  d^)  =  j5  sin'  ^, [An'-h2  Mgh  cos  6^ - E] 

+  (G- An  cos  e^f 


MOTION   OF  A  TOP 


so  that /(cos  ^„)  is  negative.    We  easily  find,  from  equation  (143), 
that 

f{i)  =  {G~An)\ 

so  that /(I)  is  positive. 

Again,  f{-l)  =  {G+.m)\ 

so  that/(—  1)  is  positive,  and 

wliich  is  negative.    Thus  we  have  seen  that 

when  cos^  =  +  oo,  /{cos  6)  is  — 

when  cos^=l,  f{cosd)  is  + 

when  cos  6=  cos  d^,  /(cos  ^)  is  — 

when  cos^=  — 1,  /(cos^)is+. 

Thus  the  three  roots  of  the  cubic /(cos  ^)=  0  lie  as  follows: 

a  root  6=6^  between  cos  6—1  and  cos  6  =  cos  0^ ; 
a  root  d=  6„  between  cos  6  =  cos  6^  and  cos  6=  —  1  ; 
a  root  for  which  cos  6  is  numerically  greater  than  unity,  giving 
no  real  value  for  6. 

We  see,  therefore,  that 
the  only  pomts  at  which 


d0 

dt 


can  vanish  are  6=6^, 


and  6  =  9^.     Moreover,  at 

these  points  —  vanishes, 
^  dt 

and  as  there  are  not  co- 
incident   roots    at    either 

.  ,  dd 
pomt  -— 
^  dt 


chanafes  sign  on 


reaching  these  points,  so  that  0  can  range  only  between  the  values 
e^  and  6'„. 

Thus  the  axis  of  the  top  oscillates  between  the  two  cones  6=  6^ 

and  e  =  e.^. 


314  MOTION    OF  KIGID   BODIES 

255.  Let  us  find  what  is  the  least  angular  momentum  wliich  the 
top  must  have  so  as  to  spui  without  falling  over.  To  do  tliis,  we 
may  assume  that  falling  will  occur  if  ever  6  exceeds  a  certain  limit 
6^,  either  through  the  peg  sHpping  or  through  its  side  touching 
the  ground.  The  condition  that  the  top  shall  not  fall  over  is  that 
^2  must  be  less  than  0^,  and  hence  that  /(cos  ^3)  must  be  positive. 
Thus  the  values  of  U,  G,  and  H  must  be  such  that 

B  sin'  0, {Aa^  +  2  Mgh  cosd^- E)  +  (G -An  cos  0,)- 

is  positive. 

Suppose  that  the  top  is  started  at  an  inclination  0^,  to  the  ver- 
tical, having  no  motion  except  one  of  rotation  II  about  its  axis. 
We  then  have,  from  equations  (140)  and  (141), 

U  =  Ail^  +  2  3Igh  cos  0^, 

G  =  An  COS  0,. 
Thus 

/(COS  0,)  =  B  sin'  0,  (An''  +  2  3Igh  cos0^-E)  +  {G-  An  cos  0,)^ 
=  Bsm'0s-2  Mgh  (cos  0^  -  cos  0^)  +  A^n''  (cos  0^  -  cos  0^f 
=  (cos ^3  -  cos 0^)  [2  3IghB  sm'  0^  +  A^n^  (cos  0^  -  cos  0^)]. 

(144) 

Since  the  top  is  necessarily  started  in  a  position  in  which  it  can 
spin,  the  value  of  cos  0^  —  cos  ^„  is  necessarily  negative.  Thus  in 
order  that  /(cos  0^)  may  be  positive,  we  must  have 

Am''  (cos  0^  -  cos  0,)  -  2  MghB  sin'  0,  (145) 

„  2  2  3fghB  sin'  0^  ,,  .  ., 

positive,  or  n^  > -—^ '—■  (146 

^  A^  (cos  0Q  —  cos  ^3) 

We  notice  that  if  A  is  very  small,  the  value  of  n  required  to 
keep  the  top  from  falling  is  very  large.  It  is  therefore  very  hard 
to  spin  a  top  of  small  cross  section,  such  as  a  lead  pencil  or  a 
pointed  wire. 

If  we  can  choose  the  angle  at  which  we  start  the  top,  cos  0^  is 
at  our  disposal.    We  see  that  the  necessary  value  for  H  is  least 


MOTION  OF  A  TOP  315 

when  cos  d^  is  a  maximum,  i.e.  when  tlie  top  is  started  vertical. 
In  this  case  the  top  will  spin  if 

^,^  2  3IffhB  sin' e, 
A' a -cose,)' 

..                              ^^  ^  2  MgJiB (1+ cos  dz) 
or  if  il^  > ^— ; '-  • 

A  ■' 

256.  In  general,  for  a  top  started  vertically  and  with  no  velocity 
except  one  of  pure  rotation  about  its  axis,  we  find,  on  putting 
cos  0^=  1  in  equation  (144), 

/  (cos  0)  =  {1-  cos  ef  [A'a""  -  2  MghB (1  +  cos  d)]. 
The  roots  of  the  equation /(cos  6)  =  0  are 

COS  ^   =  +    1,     +  1,     ; 1. 

2  MghB 

T   .             ■.           o"       ^MghB 
Let  us  write         12-  = ^ — -  > 

0  J2 

then  when  11^  =  Hq,  the  roots  are 

COS  (9  =  +  1,  +1,  +1. 

When  Q}  >  n„  the  third  root  is  greater  than  unity,  and  when 
fi^  <  Hg  the  third  root  is  less  than  unity,  say  cos  0  =  cos  0, 
where  0  is  a  real  angle,  given  by 

008©  =  - ; 1  =  1 ~rir-.  (147) 

2  3IghB  2ni  ^       ' 

Thus,  as  long  as  D.'  >  fll  the  oscillations  are  confined  within 
the  coincident  limits  ^  =  0  and  ^  =  0,  so  that  the  top  remains 
vertical,  but  as  soon  as  we  have  fl'  <  ill  the  oscillations  are 
between  the  hmits  ^  =  0  and  0  =  S. 

Suppose  we  start  a  top  with  angular  A'elocity  ft  greater  than  ftj,, 
so  that  at  first  its  axis  is  vertical  and  the  only  motion  of  the  top 
is  one  of  rotation  about  its  axis.  Then  the  real  roots  for  0  are 
0,  0 ;  there  is  therefore  no  range  of  oscillations,  and  the  axis  of 


316  MOTION  OF  RIGID  BODIES 

the  top  remains  strictly  vertical,  —  in  cumuion  language,  the  top 
is  "  asleep." 

If  the  conditions  were  the  ideal  conditions  supposed,  this  motion 
would  continue  forever,  but  in  nature  such  ideal  conditions  can- 
not exist.  The  region  of  contact  between  the  peg  and  the  surface 
on  which  it  spins  is  not  strictly  a  point,  but  a  small  circle  or 
ellipse,  on  account  of  the  small  compression  which  takes  place  at 
the  point  of  contact.  By  making  the  peg  of  hard  steel  and  spin- 
ning on  a  hard  surface,  this  region  is  very  small,  but  is  still  of 
finite  dimensions.  The  consequence  is  that  the  reactions  on  the 
peg  do  not  all  meet  the  axis.  There  is  a  small  frictional  couple 
resisting  the  rotation  of  the  top,  and  D,  gradually  decreases. 

When  XI  has  so  far  decreased  as  to  be  less  than  n^,  the  ranges 
of  oscillation  are  ^  =  0  and  ^  =  0.  The  top  is  no  longer  asleep, 
but  is  now  wobbling  through  an  angle  ©.  As  fl  continues  to 
decrease,  0  continually  increases,  as  is  clear  from  equation  (147), 
and  finally  S  reaches  so  large  a  value  that  the  top  rolls  against 
the  ground  and  so  falls  over. 

257.  The  interest  of  these  results  will  perhaps  be  enhanced  by  exam- 
ining the  form  they  assume  when  the  top  is  of  a  very  simple  kind. 
Let  us  suppose  that  a  top  is  formed  by  running 
a  pin  through  the  center  of  a  uniform  disk  of 
mass  M,  radius  a.  Let  the  length  of  the  pin 
which   protrudes   through  the  disk  on  its  lower 

side    be    h,    and    let    the    mass    of    the    pin    be 
Fig  ir>2  . 

neglected   in  comparison  with  that  of  the  disk. 

The  h  is  the  same  as  the  h   of  our  previous   analysis.    The  values  of 

A  and  B  are 


so  that  O 


2  _  4  Mgh  B  _4(jh 


When  spinning  at  the  critical  velocity  O,,  at  which  wobbling  sets  in,  the 
velocity  of  a  point  on  the  rim  is  Q^/i  or  2  ^f]h.  Thus  wobbling  begins  when 
the  velocity  of  a  point  on  the  rim  is  reduced  to  2  v^,  a  velocity  which 
depends  only  on  the  height  of  the  disk  and  not  on  its  radius.  We  see  that 
the  lower  the  disk  is,  the  slower  it  can  spin  with9ut  wobbling.  If  we  take 
h  —  2  inches,  we  find  that  wobbling  begins  when  the  rim  velocity  is  about 
4.7  feet  a  second. 


EXAMPLES  317 

The  rim  will  touch  the  ground  -when  the  range  of  wobbling  is  given  by 

tan©?=-,  and  after  this  the  tojj  will  roll  on  the  ground.    If  we  take 

a  =  G  inches  and  Ji  —  2  inches,  as  before,  this  gives  tan©=  J,  so  that 

cos©  = and  — '-^ ^  =  .100, approximately.    Thus  Q=— Q„  rouerhlv. 

VlO  ^io  20    "         ^     -^ 

Thus  such  a  top  will  "  sleep  "  until  its  rim  velocity  is  reduced  to  4.7  feet 
a  second.  After  this  it  will  wobble,  and  as  soon  as  its  rim  velocity  is 
reduced  to  about  4.5  feet  a  second  the  top  will  begin  to  roll  on  the  ground. 
For  an  ordinary  small,  pear-shaped  peg  top  we  may  take  roughly  h  —  \\ 
inches,  and  the  radii  of  gyi-ation  abouk  axes  through  the  point  of  contact 
as  I  inch  and  2  inches.    Thus  in  inches 

A  =  i?5  M,         i?  =  4  M, 

,      4  Mf/h  B      2048 

Q  n  = '- = a. 

A'^  27 

Taking  g  =  386  inches  per  second  per  second,  this  gives  fi„  =  170  revolu- 
tions per  second.  If  thrown  from  a  string  of  which  the  end  is  coiled  round 
the  top  in  circles  of  radius  1  inch,  the  string  must  be  withdrawn  with  a 
velocity  of  about  60  miles  an  hour  relative  to  the  top  to  set  up  the  required 
angular  velocity. 

GENERAL   EXAMPLES 

1.  A  fly  wheel  whose  moment  of  inertia  is  /  has  a  string  wound  round 
its  axle  of  radius  h.  A  tension  equal  to  a  weight  iv  is  applied  to  the  string 
for  1  second.  What  is  the  angular  velocity  of  the  fly  wheel  at  the  end  of 
1  second? 

2.  A  fleet  of  total  displacement  200,000  tons  steams  from  east  to  west 
along  the  equator,  covering  20  minutes  of  longitude  per  hour.  Regarding 
the  earth  as  a  homogeneous  sphere  of  mass  6  x  10-^  tons,  find  the  change 
in  the  earth's  angular  velocity  produced  by  the  motion  of  the  fleet.  Show 
that  the  day  is  lengthened  by,  roughly,  16  x  lO"^'*  seconds. 

3.  The  earth's  mass  is  6  x  lO-^  tons,  and  icebergs  and  melted  snow 
weighing  10^''  tons  move  from  the  north  pole  to  latitude  45  degrees.  Find 
the  change  in  the  lengtli  of  the  day. 

4.  A  train  of  mass  m  runs  due  north  at  CO  miles  an  liour.  Slu)\v  tliat 
there  must  be  a  pressure  between  the  eastern  rail  and  the  flanges  of  tJie 
wheel  in  consequence  of  the  earth's  rotation,  and  find  the  amount  of 
the  pressure. 

5.  A  thin  layer  of  dust,  thickness  //  feet,  is  formed  on  the  earth  by 
the  fall  of  meteors,  reaching  the  earth  from  all  directions.    Show  that  the 


318  MOTION   OF   IIIGID   BODIES 

change  in  the  length  of  a  day  is  about  — —  of  a  day,  where  a  is  tlie  radius 

of  the  earth  in  feet  and  D,  p  are  the  density  of  tlie  earth  aud  the  meteoric 
dust  respectively. 

6.  Two  masses  M  and  m  suspended  from  a  wheel  and  axle  of  radii 
a,  /;  do  not  balance.    Show  that  the  acceleration  of  M  is 

Ma  —  mh 


Ma^  +  mb'^  +  1 
where  I  is  the  moment  of  inertia  of  the  machine  about  its  axis. 

7.  A  uniform  cylinder  has  coiled  round  its  central  section  a  light,  pei-- 
fectly  flexible,  inextensible  string.  One  end  of  the  string  is  attached  to  a 
fixed  point,  and  the  cylinder  is  allowed  to  fall.  Show  that  it  will  fall  with 
acceleration  f  g. 

8.  Two  equal  uniform  rods  of  length  2  a,  loosely  joined  at  one  extremity, 

a  V2 
are  placed  synimetrically  upon  a  fixed  sphere  of  radius  — ; —  and  raised 

o 

into  a  horizontal  position  so  that  the  hinge  is  touching  the  sphere.  They 
are  then  allowed  to  descend.  Show  that  when  they  are  first  at  rest  they 
are  inclined  at  an  angle  cos"i|^  to  the  horizontal,  that  the  pressure  on  the 
sphere  at  each  point  of  contact  is  one  quarter  of  the  weight  of  a  rod,  and 
that  there  is  no  strain  at  the  joint. 

9.  A  rod  rests  with  one  extremity  on  a  smooth  horizontal  plane  and  the 
other  on  a  smooth  vertical  wall,  the  rod  being  inclined  at  an  angle  a  to 
the  horizon.  If  it  is  allowed  to  sliji  down,  show  that  it  will  separate  from 
the  wall  when  its  inclination  to  the  horizontal  is  sin~'  (|  sin  a). 

10.  If  the  sun  gradually  contracts  in  such  a  way  as  always  to  remain 
similar  to  itself  in  constitution  and  form,  show  that  when  every  radius  has 
contracted  an  nth  part  of  its  length,  where  n  is  large,  the  angular  velocity 

will  have  increased  to  ( 1  +  -|  times  its  former  value.    Examine  the  change 

in  the  kinetic  energy  of  rotation. 

11.  An  elastic  band  of  natural  length  2  7ra,  mass  m,  modulus  X,  rests 
against  a  rough  wheel  of  radius  a  in  a  horizontal  plane.  The  string  is  held 
against  the  circumference  of  the  wheel,  which  is  made  to  rotate  with  angu- 
lar velocity  ft.    If  the  string  is  left  to  itself,  show  that  it  will  expand,  and 

that  when  its  radius  is  r  its  angular  velocity  will  be  — ;—  ,  and  that  its  radial 
velocity  will  be 

L  r'^  ma         J 


EXAMPLES  319 

12.  A  uniform  triangular  disk  ABC  is  so  supported  that  it  can  oscillate 
in  its  own  plane,  which  is  vertical,  about  A.  Show  that  the  length  of  the 
simple  equivalent  penduhim  is 

1     :]  (1/^  +  c^)  -  a- 

13.  A  spherical  hollow  of  radius  n  is  made  in  a  cube  of  glass  of  mass  M, 
and  a  particle  of  mass  m  is  placed  inside.  The  cube  is  then  projected  with 
velocity  F  on  a  smooth  horizontal  plane.  If  the  particle  just  gets  round 
the  sphere,  remaining  in  contact  with  it  all  the  way,  show  that 

I  2  ^  y  fl^  +  4  ag  —  ■ 

14.  Three  equal  particles  are  attached  to  the  ends  and  middle  point  of  a 
rod  of  negligible  mass,  and  one  of  the  end  particles  is  struck  by  a  blow  at 
right  angles  to  the  rod.  Show  that  the  velocities  of  the  particles  at  starting 
are  in  the  ratio  -    ^    . 

15.  A  rough  horizontal  cylinder  of  mass  M  and  radius  a  is  free  to  turn 
about  its  axis.  Round  it  is  coiled  a  string,  to  the  free  extremity  of  which 
is  attached  a  chain  of  mass  m  and  length  /.  The  chain  is  gathered  close 
up  and  then  let  go.  Show  that  if  9  is  the  angle  through  which  the  cylinder 
has  turned  after  a  time  t  before  the  chain  is  fully  stretched,  then 


a.= 


Mad  =  —  (-r//2  -00 

16.  A  uniform  flat  circular  disk  is  projected  on  a  rough  horizontal  table, 
the  friction  on  any  element  moving  with  velocity  V  being  cV^  x  (mass  of 
element),  in  a  direction  opposite  to  that  of  V.  Find  the  path  of  the  center 
of  the  disk. 


CHAPTER  XII 
GENERALIZED  COORDINATES 

258.  So  far  we  have  dealt  with  the  luechanics  (dynamics  and 
statics)  of  material  bodies  on  the  supposition  that  these  bodies 
consist  of  innumerable  small  particles  wliich,  in  the  case  of  a  rigid 
body,  are  held  firmly  in  position  and  serve  the  purpose  of  trans- 
mitting force  from  one  part  of  the  body  to  another. 

259.  Even  when  dealing  with  rigid  bodies  this  conception  of  tlie 
structure  of  matter  has  not  led  to  entirely  consistent  results.  For 
instance,  we  have  found  that  after  an  impact  between  two  im- 
perfectly elastic  bodies,  or  after  sliding  between  two  imperfectly 
smooth  bodies,  a  certain  amount  of  energy  disappears  from  view, 
and  we  have  had  to  suppose  that  this  energy  is  used  m  startmg 
small  motions,  relative  to  one  another,  of  the  ultimate  particles  of 
which  the  bodies  are  composed.  In  other  words,  after  an  impact 
or  sliding  has  taken  place,  a  rigid  body  can  no  longer  be  supposed 
to  satisfy  the  conditions  postulated  for  a  rigid  body. 

When  dealing  with  bodies  which  are  obviously  not  rigid  the  case 
is  worse.  Here  the  conceptions  which  we  have  introduced  into  the 
study  of  rigid  bodies  do  not  help  at  all,  and  very  little  progress  is 
possible  without  introducing  some  other  conceptions  to  replace  these. 

260.  There  are  two  ways  of  proceeding  at  this  stage.  We  may 
introduce  new  conceptions  which  seem  plausible,  and  in  this  way 
try  to  form  a  picture  of  the  structure  of  the  matter  with  which  we 
are  dealing.  We  cannot  be  certain  that  the  results  obtained  in  this 
way  will  be  true,  for  we  can  never  be  sure  that  our  conceptions  of 
the  nature  of  the  ultimate  structure  of  matter  are  accurate.  But 
it  may  be  worth  trying  what  results  are  obtained  by  introducing  a 
set  of  provisional  conceptions  as  to  the  structure  of  matter.  If 
these  results  are  in  agreement  with  the  phenomena  observed  in 

320 


CxEXEPvALIZED  COORDINATES  321 

nature,  the  probability  that  ovir  provisional  conceptions  are  near  to 
the  truth  is  strengthened.  If,  on  the  contrary,  the  results  obtained 
are  not  found  to  agree  with  what  is  observed  in  nature,  the  pro- 
visional conceptions  from  which  these  results  have  been  deduced 
must  be  either  modified  or  withdrawn. 

Different  sets  of  conceptions  as  to  the  structure  of  the  matter 
dealt  with  will  lead  to  different  branches  of  mathematical  physics. 
As  instances  of  such  branches  of  mathematical  physics  may  be 
mentioned  the  theory  of  elastic  solids  which  is  based  upon  certain 
provisional  conceptions  as  to  the  behavior  of  the  particles  of 
which  solid  bodies  are  composed,  and  the  kinetic  theory  of  gases 
which  is  based  upon  certain  provisional  conceptions  as  to  the 
behavior  of  the  particles  of  a  gas.  The  tracmg  out  of  the  conse- 
quences of  different  sets  of  provisional  conceptions  as  to  the  struc- 
ture of  matter  cannot,  however,  be  regarded  as  coming  within  the 
scope  of  a  book  such  as  the  present  one. 

261.  There  is,  however,  an  alternative  way  of  proceeding.  We 
have  taken  Newton's  laws  of  motion  as  the  material  supplied  by 
experimental  science  for  theoretical  science  to  work  upon.  The 
truth  of  these  laws  as  applied  to  the  ultimate  particles  of  the 
material  universe  is  by  no  means  certain,  because  we  cannot  obtain 
the  ultimate  particles  to  experiment  upon.  Suppose,  however,  that 
we  examine  whether  any  further  progress  can  be  made  in  the 
study  of  mechanics  without  introducing  any  hypothesis  beyond 
the  smgle  one  (admittedly  uncertain)  that  Newton's  laws  apply  to 
the  ultimate  particles.  If  w^e  can  make  progress  in  this  direction, 
the  results  obtained  will  of  course  apply  to  all  further  extensions 
of  mechanics,  whether  or  not  additional  hypotheses  are  introduced 
as  to  the  nature  and  arrangement  of  the  ultimate  particles. 

262.  The  standpoint  from  wliicli  we  are  regarding  the  matter 
cau,  perhaps,  be  explained  by  an  analogy,  first  suggested  by  Pro- 
fessor Clerk  Maxwell.  Suppose  that  we  have  a  complicated  ma- 
chine in  a  closed  room,  and  that  the  only  connection  between 
this  machine  and  the  outer  world  is  by  means  of  a  number  of 
ropes  which  hang  through  holes  in  the  floor  into  the  room  beneath. 


322  GENERALIZED   COORDINATES 

A  man  introduced  into  the  room  beneath  will  have  no  opportunity 
of  inspecting  the  machinery  above,  but  he  can  manipulate  it  to  a 
certain  extent  by  pulling  the  different  ropes.  If  on  pulling  one 
rope  he  finds  that  the  others  are  set  into  motion,  he  will  under- 
stand that  the  different  ropes  must  be  connected  above  by  some 
kind  of  mechanism,  but  will  not  be  able  to  discover  the  exact 
nature  of  the  mechanism. 

This  concealed  mechanism  may  be  supposed  to  represent  tliose 
parts  of  the  mechanism  of  the  universe  which  are  hidden  from  our 
view,  while  the  ropes  represent  those  parts  which  we  can  manipu- 
late. In  nature,  there  are  certain  acts  which  we  can  perform,  cor- 
responding to  the  pulling  of  the  ropes  in  our  analogy,  and  we  see 
that  these  are  followed  by  certain  consequences,  analogous  to  the 
motion  of  the  other  ropes ;  l)ut  the  ultimate  mechanism  by  which 
the  cause  produces  the  effect  remains  entirely  unknown  to  us. 
For  mstance,  if  we  press  the  key  of  an  electric  circuit,  we  may 
find  that  the  needle  of  a  distant  galvanometer  is  moved,  but  the 
mechanical  processes  which  transmit  the  action  through  the  wires 
of  the  circuit  and  through  the  ether  surrounding  the  galvanometer 
needle  remain  unknown. 

263.  Now  suppose  that  the  imaginary  man  is  at  liberty  to  handle 
the  ropes  and  that  he  wishes  to  study  the  connection  between  them. 
He  may  begin  by  conjecturing  that  the  connecting  mechanism  in 
the  room  above  consists  of  arrangements  of,  say,  levers,  pulleys,  and 
cogwheels,  and  he  may  work  out  for  himself  the  manner  in  which 
the  ropes  ought  to  move  if  his  conjectures  are  correct.  This  proced- 
ure would  be  analogous  to  that  we  have  described  in  §  260;  it  is 
not  the  procedure  we  are  going  to  follow  here. 

On  the  other  hand,  without  any  conjecture  at  all  as  to  the 
nature  of  the  mechanism  above,  the  man  will  know  that  certain 
laws  will  govern  any  manipulation  of  the  ropes,  if  the  ropes  are 
connected  by  mechanism  of  any  kind  whatever,  such  that  each 
particle  obeys  Newton's  laws  of  motion. 

To  explain  this,  let  us  take  the  simplest  case,  and  suppose  that 
there  are  two  ropes  only  and  that  when  A  is  pulled  down  half  an 


HAMILTON'S  PRINCIPLE  323 

inch,  then  B  invariably  rises  through  two  inches.  The  mechanism 
may  be  a  lever,  an  arrangement  of  pulleys,  or  clockwork.  Jiut 
whether  it  is  any  one  of  these,  or  something  entirely  different  from 
any  of  them,  it  will  be  known  that  the  motion  of  rope  A  down- 
wards can  be  restrained  by  exerting  on  rope  B  a  force  equal  to  half 
of  that  applied  to  A.  This  fact  follows  from  the  priuciple  of  virtual 
work,  quite  apart  from  any  conjecture  as  to  the  nature  of  the 
hidden  mechanism.  Now  the  question  before  us  is  as  follows : 
Can  we,  without  any  knowledge  of  the  Iddden  mechanism,  discover 
what  motion  of  the  ropes  will  ensue,  if  they  are  started  m  any 
given  way.  And  the  answer  is  that  we  can,  provided  we  know 
the  amount  of  energy  involved  in  a  motion  of  any  kind,  —  i.e.  pro- 
vided we  know  the  kinetic  energy  of  every  motion,  and  also  the 
potential  energy  of  every  configuration. 

So  also,  to  pass  from  analogies  to  realities,  we  can,  without  any 
knowledge  of  the  rdtimate  mechanism  of  the  universe,  discover 
what  motion  will  ensue  from  any  initial  conditions,  provided  tliat 
we  know  the  kinetic  and  potential  energies  of  all  configurations 
of  the  portion  of  the  imiverse  with  which  we  are  dealing. 

Hamilton's  Principle 

264.  Let  us  suppose  that  any  single  particle  of  a  material  sys- 
tem has  at  any  instant  coordmates  x^,  i/i,z^,  its  mass  being  7?^J,and 
that  it  is  acted  upon  by  forces  of  which  the  resultant  has  compo- 
nents X^,  1\,  Z^.  Let  the  velocity  of  this  particle  have  components 
%,  t'j,  %i\,  so  that  7  , 

v.,  =  — i  >  etc. 
'      dt 

Then,  if  the  motion  of  this  particle  is  governed  by  Newton's  laws, 
we  shall  have  , 

^^=X^,  (148) 

m,^=Fi,  (149) 

mf-^=Z,.  (150) 

at 


324  GENERALIZED   COORDINATES 

Let  us  compare  this  motion  with  a  slightly  different  motion  in 
wliich  Newton's  laws  are  not  obeyed.  Li  this  second  motion  let  the 
coordinates  of  ni^,  at  the  instant  at  wliichthey  are  x^,  y^,  z^  in  the 
actual  motion,  be  supposed  to  be  x[,  y[,  z[,  and  let  the  components 
of  velocity  at  this  instant  be  u[,  v[,  iv[,  so  that 

dx' 
u',  =  -~,  etc. 
'       dt 

Let  us  agree  that  the  modified  motion  is  to  differ  so  slightly 
from  the  actual,  that  any  quantity  such  as  x[— x^,  u[—  u^,  which 
measures  part  of  this  difference,  may  be  treated  as  a  small  quan- 
tity. Let  us  denote  x[  —  x^  by  hx^,  and  use  a  similar  notation  for 
the  other  differences. 

Multiply  equations  (148),  (149),  (150),  which  are  true  at  every 
instant,  by  hx^,  hy^,  hz.^,  and  add.    AVe  obtain 

di(,  5.      ,         dv.  ^      ,         div,  5, 
m,  —r-  ox,  +  m,  — -^  by,  +  tn,  — -  oz, 
^  dt    ^^         '  dt     ^^         ^  dt       ^ 

=  X,hx^  +  l\hy^+  Z,Sz,.  (151) 

Now  'I^8x,  =  f^(uM)-n4t^^"'^^ 

=  dt^'^^^-"^dt^^-''^ 

d 

dt^   ^     ''         '     ' 


Hence 


dii,  5,      ,        dr,  ^      ,        dio,  5, 


=  m 


u^8x^  +  v^Sy^  +  v\8zj)  —  (?^i3«i  +  t\8vi  +  tv^Sw^) 
=  A\Bx^  +  y^Sy^  +  Z,8z,,  (152) 


d 
dt^' 


by  equation  (151). 


HAMILTON'S  PEIXCIPLE  325 

An  equation  of  this  kind  is  true  for  each  particle  of  the  system 
and  at  every  instant  of  the  motion.  It  is  moreover  true  whatever 
the  displaced  motion  may  be.  On  summing  this  equation  for  all 
particles  we  obtain 

\^  cl  ~\ 

=  ^{A\8x,  +  l\8i/,  +  Z,8z,).  (153) 

Now  let  T  denote  the  kinetic  energy  of  the  motion,  so  that 

Then  8T  =  ^Vm,{t([^-  ir,  +  r[ '  -  r  j  +  iv[ = -  i/r^ ). 

Now        i(['^  —  2q  =  (w^  +  8uJ~  —  vi  =  2u^8u^, 

if  we  neglect  the  small  quantity  of  the  second  order  {8iiiY,  ^^  ^^^^^ 

>ve  have 

BT  =  ^ni^{u^8a^  +  v^8i\  +  u\8w^). 

265.  Assummg  for  the  moment  that  the  system  of  forces  is 
conservative,  let  IF  denote  the  potential  energy  of  the  system  at  the 
instant  under  consideration,  and  W  that  of  the  imaginary  system 
in  the  slightly  displaced  configuration.    Then,  by  §  118,  we  have 

8W=W'-W 

=  (work  done  in  movmg  system  from  actual 

to  displaced  configuration) 
=  -  2;  (XM+1\8>/,  +Z^z;).  (154) 

Substituting  into  equation  (153)  for  the  expressions  which  have 
been  foimd  to  be  equal  to  8T  and  8W,  we  find  that  this  equation 
reduces  to  the  simpler  form 

^m,  I  (u,8.:,  +  r,8i/,  +  .-,§.,)  -8T  =  -  8JV, 
or  again. 


326  ge:^ieralized  coordinates 

This  equation  is  true  at  every  instant  of  the  motion.  Let  us 
integrate  it  between  any  two  instants  of  the  motion,  say  from 
t  =  t^  to  t  =  t^-    We  obtain 

^y^m,{u,B.r^  +  r,8ij^  +  w,Sz,)Y^^  =  C'^'SiT -ir)dt.        (155) 

The  displaced  motion  has  so  far  been  subject  to  no  restrictions 
except  that  the  difference  between  it  and  the  actual  motion  must 
always  remain  small.  Let  us  now  introduce  the  further  restriction 
that  at  times  t.^  and  t^  the  configurations  in  the  displaced  motion 
are  to  be  identical  with  those  in  the  actual  motion.  The  displaced 
motion  is  now  one  in  which  the  imaginary  system  starts  in  the 
same  configuration  as  the  actual  system  at  time  t  =  t^,  swerves 
from  the  course  of  the  actual  system  from  time  ^^  to  time  t^ 
(because  the  actual  system  obeys  Newton's  laws,  while  the  imagi- 
nary system  does  not),  and  ultimately  ends  in  the  same  position 
as  the  actual  system  at  time  t^. 

In  consequence  of  this  restriction  on  the  motion  of  the  imagi- 
nary system,  we  have  at  times  t^  and  t^, 

and  similar  relations  for  the  other  particles.    Thus 
and  equation  (155)  reduces  to  , 


r 


S{T-W)dt  =  0.  (156) 


Here  we  have  an  equation  which  depends  only  on  the  amounts 
of  the  kinetic  and  potential  energies  of  the  system,  and  not  on  the 
mechanism  of  the  system.  We  shall  find  that  from  this  single  equa- 
tion we  can  determine  the  motion  of  all  the  known  parts  of  the 
system  as  soon  as  T  and  W  are  known,  without  any  knowledge  of 
the  mechanism  of  the  unknown  parts. 


PEINCIPLE  OF  LEAST  ACTION  327 

266.  Before  proving  this,  however,  we  may  attempt  to  interjjret 
equation  (156).    Let  us  denote  T  —  Whj  L.    Then 


f\{T-W)dt=  f  'hLdt 

=  r  \l'  -  L)  dt 
=  f'L'df-  f  'l 

I. 


dt 

h 

Ldt' 


>'2 

If  we  denote  /    L  dt 


by  *S',  the  equation  becomes  8S  =  0,  or 

S'  =  S. 

Tlius  the  value  of  the  function  S  for  the  actual  motion  is  the 
same,  except  for  small  quantities  of  the  second  and  higher  orders, 
as  the  corresponding  function  S'  for  any  slightly  different  motion, 
which  begins  and  ends  with  the  same  configuration  at  the  same 
instants.  In  other  words,  the  function  S  is  either  a  maximum  or  a 
miaimum  when  the  series  of  configurations  is  that  which  actually 
occurs  in  nature. 

Principle  of  Least  Action 

267.  The  total  energy  will,  by  the  theorem  of  §  143,  remain 
constant  durmg  the  actual  motion,  say  equal  to  ^,  so  that  at  every 
instant  we  shall  have 

T  +  W  =E,       T-  W  =  L. 

In  the  slightly  varied  series  of  configurations  it  is  not  true  that 
the  total  energy  remams  constant  throughout  the  motion,  but  out 
of  the  infinite  number  of  slightly  varied  series  of  configurations 
there  will  still  be  an  infinite  number  for  which  the  conditions 
already  postulated  are  satisfied,  together  with  the  condition  that 
the  total  energy  at  every  instant  shall  have  the  value  E.    For  such 


328  GENERALIZED  COORDINATES 

a  series  we  have 

T'  +  W'  =  E,       T'  -  W  =  L'. 
Thus  we  have    L  =  2T-E,      L'  =  2T'-E, 

so  that  ^  ~  \    ^  '^^ 

=  r  \2T-  E)dt 

=  C  '2Tdt-{t-t,)E. 

Thus  if  S  is  a  maximum  or  a  mmimum,  it  follows  that 

'2Tdt 


I 


is  a  maximum  or  a  minimum.  This  integral  is  called  the  action 
of  the  motion.  We  now  see  that  of  all  possible  series  of  configura- 
tions which  bring  the  system  from  one  configuration  to  another  in 
a  given  time,  and  in  such  a  way  that  the  total  energy  has  always 
a  specified  constant  value,  that  one  which  can  be  described  by  a 
natural  system  is  the  one  on  which  the  action  is  a  maximum  or 
a  minimum.  Since  the  action  is  in  general  a  minimum,  this  prin- 
ciple is  known  as  the  principle  of  least  action. 

The  statement  of  this  principle  was  first  given  by  Maupertius 
(1690-1759),  who  did  not  deduce  it  by  mathematical  reasoning,  but 
believed  it  could  be  proved  by  theological  arguments  that  all  changes  in 
the  universe  must  take  place  so  as  to  make  the  action  a  minimum  (^Essai 
de  Cos  in  ologie ,  1751). 

NON-CONSERVATIVE    FORCES 

268.  If  the  forces  are  non-conservative,  we  may  no  longer,  as  in 
equation  (154),  replace 

by  —  8W,  and  consequently,  instead  of  equation  (156),  we  shall  have 
f  '  [8T+^(X8x  +  YSi/  +  Z8z)'^  dt  =  0.  (157) 


LAGRANGE'S  EQUATIONS  329 

Lagrange's  Equations 

269.  If  the  coordinates  x^,  y^,  z^,  etc.,  of  every  particle  of 
the  system  are  known,  we  know  not  only  the  configuration  of 
the  system  but  also  the  mechanism  by  which  the  different  parts 
of  the  system  are  connected.  It  may,  however,  be  that  we  can 
determine  the  configuration  of  the  system  by  knowing  a  smaller 
number  of  quantities  which  do  not  give  us  a  knowledge  of  the 
mechanism. 

For  instance,  in  our  former  illustration  we  imagined  two  roi:)es  to  hang 
from  an  unknown  machine,  the  ropes  being  connected  in  such  a  way  that 
a  motion  of  one  inch  in  the  one  invariably  produced  a  motion  of  two  inches 
in  the  other.  In  this  case  the  configuration  is  fully  determined  when  we 
know  the  single  coordinate  which  measures  the  position  of  the  end  of  the 
first  rope,  but  a  knowledge  of  this  coordinate  does  not  imply  a  knowledge 
of  the  mechanism  connecting  the  ropes. 

Again,  the  position  of  a  rigid  body  is,  as  we  have  seen  (§  Go),  determined 
by  the  values  of  sufficient  quantities  (six)  to  fix  the  positions  in  space  of 
three  non-collinear  particles  of  the  body,  but  a  knowledge  of  these  quanti- 
ties does  not  give  us  information  as  to  the  arrangement  of  the  particles  of 
which  the  body  is  formed. 

Let  ^p  0„,  •  ■  • ,  6^  be  a  set  of  quantities  such  that  w^hen  their 
value  is  known,  the  configuration  of  a  system  of  bodies  is  fully 
determined.  Then  the  quantities  6^,  0^,  •  •  •,  6^^  are  called  general- 
ized coordinates  of  the  system. 

270.  Let  X,  y,  z  be  the  coordinates  of  any  particle  of  the  system. 
Then  x  is  fully  determined  by  the  values  of  6^,  6.,,  ■  ■  ■,  9^^,  so  that 
it  is  a  function  of  these  quantities,  say 

^-f(^r,   ^.,    ■•■,    ^n)-  (158) 

If  the  system  is  m  motion,  all  the  quantities  which  enter  in 
equation  (158)  are  functions  of  the  time.  We  have,  on  differ- 
entiation with  respect  to  the  time, 

dt~ dd.  dt '^  te„ll  '^  "  '  ^ dd„  dt  ' 


330  GENERALIZED  COORDINATES 

dx    d6 
To  abbreviate,  let  us  denote  —r '  -r^  >  ■  •  •  hy  x,6,,-  ■  -.    Then  the 

di     dt  ^         ^ 

equation  just  obtained  may  be  written 

so  that  i;  is  a  linear  function  of  Q^,  G^,-'-,  ^„,  the  coefficients  being 
functions  oi  6.^^,  0^,  ■  ■  ■ ,  6^^. 
The  kinetic  energy, 

is  now  seen  to  be  a  quadratic  function  oi  6^,  6^,  •  ■  ■ ,  6^^,  the  coeffi- 
cients bemg  functions  of  6^,  6^,  •  •  ■ ,  ^„. 

The  potential  energy  W  depends  only  on  the  configuration  of 
the  system,  so  that  JV  is  a  function  of  ^^,  ^2>  *  *  '>  ^«  '^^J- 

Thus  the  function  L,  or  T  —  W,  is  a  function  of 

^i,  ^2>  •  •  •'  ^n'  ^1'  ^2'  •  •  •»  ^«» 

say  L  =  4>{e„  e„  . . .,  e,,  e„  e,,  ■ . .,  e,).  (leo) 

The  corresponding  function  L'  m  the  displaced  motion  is  tlie 
same  function  of 

e^  +  w^,  e.,+8e„,  ...,  etc., 

so  that 

By  Taylor's  theorem,  we  may  expand  L'  in  the  form 


or,  from  equation  (160), 


L'  =  L+^he,'^+^W,'±.  (161) 


£ 


LAGKANGE'8  EQUATIONS 

Equation  (156),  namely 

'8{T-W)dt  =  0, 
may  be  written  in  the  form 

\l'  -  L)  dt  =  0, 
and  this,  we  now  see,  may  be  replaced  by 

Now  we  have 

w,  =  e[-e,  =  ^^  (0,  +  86,)  -  ~  ^x  =  y,  m. 


331 


£ 


(162) 


dt 


so 


that  f '"'  —  Se,  dt  =  r ''  -^  4  (S^i)  ^^ 

and,  integrated  by  parts,  this  becomes 

^^hoX-  f'^'^(—)sd,dL  (163) 

Since  the  disturbed  configuration,  by  hypothesis,  coincides  with 
the  actual  configuration,  we  have  S^^  =  0  at  times  t,  and  t^.  Thus 
the  first  term  in  expression  (163)  vanishes,  and  leaves 

Equation  (162)  now  assumes  the  form 

\  dt  =  0. 


Jii      I    1 


az  _  ^  /dL\ 

cO,    dt\cej\} 


(164) 


The  limits  t,  and  t.^  are  entirely  at  our  disposal ;  the  equation  is 
true  whatever  values  we  assign  to  them.  In  other  words,  the  sum 
of  a  number  of  small  differentials  vanishes,  no  matter  how  many 


GENERALIZED  COOEDINATES 


of  them  are  included  iii  the  sum.    It  follows  that  each  term  of  the 
sum  must  vanish.    Thus  we  must  have 


se. 


dL 


dt  \dd. 


=  0 


(165) 


at  every  instant. 

271.  At  this  point  we  have  to  consider  two  alternatives.  It  may 
be  that  whatever  values  are  assigned  to  hd^,  B6„,  ■  ■  ■ ,  86^^,  the  new 
configuration,  specified  by  coordinates 

will  be  a  possible  configuration ;  that  is  to  say,  will  be  one  which 
the  system  can  assume  without  violating  the  constraints  imposed 
by  the  mechanism  of  the  system.  In  this  case  the  system  is  said 
to  have  n  degrees  of  freedom. 

If  the  system  has  oi  degrees  of  freedom,  equation  (165)  is  true 
for  all  values  of  80^,  80^,  ■  •  • ,  S0^^.    For  instance,  it  is  true  if  we  take 

80,  =  e,  80,=^  80,=^.-.^  80,^  =  0, 

where  e  is  any  small  quantit}'.    In  this  case  we  must  have 

=  0, 


and  therefore 


az  _  ^  /dL 
30,      dt  \d0^ 

d^/dL 
dt  \d0 


-?4  =  0. 


A  similar  equation  will,  of  course,  hold  for  each  of  the  coordinates 
0,,  0^,  •■  ■ ,  0^^.  These  equations  are  known  as  Lagrange's  equa- 
tions. There  are  n  equations  between  the  n  unknown  quantities 
0,,  0,,,  •  ■  ■,  0„  and  their  differential  coefficients  with  respect  to  the 
time.  Thus  they  enable  us  to  find  the  way  in  which  0,,  0^,  •  •  -,  0„ 
change  with  the  time.  To  use  the  equations  we  require  a  knowl- 
edge only  of  the  function  L,  and  therefore  only  of  the  kinetic  and 
potential  energies  of  the  S3^stem ;  we  do  not  need  a  knowledge  of 
the  internal  mechanism  of  the  system.  Thus  the  prol)lem  proposed 
in  §  263  is  solved,  if  we  can  solve  Lagrange's  equations. 


LAGEAXGE'S  EQUATIONS 


333 


ILLUSTRATIVE  EXAMPLE 

Common  pendulum.  As  a  simple  example  of  the  use  of  Lagrange's  equations, 
let  us  consider  the  problem  of  the  motion  of  the  common  pendulum.  A  rigid 
body  is  constrained  to  move  so  that  one  jjoint  O  remains  fixed,  while  the  line 
OG  joining  O  to  the  center  of  gravity  moves  in  a  vertical  plane.  Let  6  be  the 
inclination  of  OG  to  the  vertical;  then  the  position  of  the  system  is  entirely 
fixed  as  soon  as  the  value  of  0  is  known.  The  kinetic  and  potential  energies  are, 
in  the  notation  of  §  245, 


so  that 


r  =  i  Mk-^e'^,  W  =  Mgh  (1  -  cos  6), 
L-\  Mm'^  -  Mgh  (1  -  cos  0). 


dL 


Thus  — r  =  Mk:-0,  and  Lagrange's  equation 

de 


dt\d0 


cL 


becomes 


Mk^  —  =  -  Mgh  sin  0, 


the  same  equation  as  was  obtained  in  §  245,  and  from  this  the 

Fig    15.3 
motion  can  be  deduced. 

We  notice,  however,  that  Lagrange's  method  shows  that  the  motion  is  inde- 
pendent of  the  method  of  suspension  of  the  pendulum,  provided  only  that  it  is 
constrained  to  move  in  the  way  described.  For  instance,  the  result  is  true  if 
there  is  no  pivot  at  all  at  0,  the  constraints  being  imposed  by  a  suspension  of 
strings. 


272.  Let  us  now  consider  the  second  alternative  to  that  exam- 
ined in  §  271.  It  may  be  that  if  we  assign  arbitrary  values  to 
W^,  86^,  •  ■  •,  86^^,  the  new  configuration  obtained  is  not  in  every 
case  a  possible  one.  It  may  be  that  there  are  certain  relations 
which  must  be  satisfied,  in  order  that  the  constramts  imposed  by 
the  mechanism  may  not  be  violated. 

For  instance,  in  the  illustration  already  employed,  let  there  be  two 
ropes  hanging  from  a  ceiling  of  a  room,  such  that  on  pulling  one  down  one 
inch  the  mechanism  compels  the  second  to  rise  two  inches.  Let  0^,  ^., 
denote  the  lengths  of  ropes  below  the  ceiling.  Then  a  displacement  in 
which  d0^  =  y'j  inch,  5^2  ==  sV  inch,  is  not  a  possible  displacement ;  such 
a  displacement  is  not  permitted  by  the  mechanism  above.  "We  must  always 
have  5^^,  5^2  connected  by  the  relation 


5^1  +  .1  5612  =  0. 


334 


GENERALIZED  COOEDmATES 


In  general  let  us  suppose  that  we  have  certain  relations  imposed 
by  the  mechanism,  these  being  of  the  form 


Then  equation  (165),  namely 

n 


dt\ddj      ^ 


S(9j=0, 


(166) 
(167)... 


(168) 


is  true  only  if  hd^,  hd.^,  ■  ■  ■,  S^„  satisfy  relations  (166),  (167),  •  •  ■. 

Eor  a  possible  displacement,  however,  hd^,  86^,  •  •  •,  86^  will  be 
such  that  equations  (166),  (167)  .  .  •,  and  (168)  are  all  true.  Let 
us  multiply  by  X,  /n,  •  • .  and  unity,  and  add,  X,  /x,  • .  •  being  quan- 
tities as  yet  undetermined  —  undetermined  multipliers,  we  may  call 
them.    Then  we  have  the  equation 


[dt  \ddj 


dL 


+  \a^  +  lxh^-{- 


86, 


d  /cL\      dL     , 

dt  \dej    sd.^ 


86. 


+ 


d  /dL\      dL      ^ 


8^,  =  0.(169) 


Tlie  quantities  86^,  86„,  ■  ■  ■ ,  86^^  are  not  at  our  disposal.  If,  how- 
ever, the  relations  of  the  type  (166)  are  m  in  number,  we  may  say 
that  of  the  quantities  86 ^,  86^,  ■  ■  ■ ,  86^^  all  except  m  are  at  our  dis- 
posal, and  after  arbitrary  values  have  been  assigned  to  n  —  m  of 
these  quantities,  the  remaining  m  quantities  must  be  obtained  by 
solving  equations  (166),  (167)  •  •  • .  The  configuration  obtained  in 
this  way  must  necessarily  be  a  possible  one. 

Let  us  assign  arbitrary  values  to 


S^,„-M>  ^^„ 


•,  ^^,. 


LAGRANGE'S  EQUATIONS  335 

and  then  find  the  values  of  86^,  86^,  ■  ■  -,  S^„,  from  equations  (16G), 
(167)  •  •  •.  Let  us,  moreover,  choose  the  7n  undetermined  multi- 
pliers X,  /i,  •  •  •  so  that  they  satisfy  the  m  equations 


(l)~l+^"'+''''+---  =  °'         <"°' 


±(dL\      dL 
dt\dd 


d  /dL\       cL      ^  , 

the  suffixes  ranging  from  1  to  m.    Then  equation  (169)  reduces  to 
d  /  dL  \        dL 


X 


dt  \dd      /      ^Q, 


+  ^«m+:  +  ^?Wi  + 


m+1 


s^,„.,  =  o. 


Inasmuch  as  S^,„+i,  5^„^^2>  •  '  "»  ^^«  ^^'^  all  arbitrary,  we  may  take 


and  obtain 

d  /  dL  \        dL 


di  \dd,n+J      cd 


+  Xfr,„^i  +  ^&„,^i+-..  =  0; 


and  similarly  we  may  obtain  the  same  equation  for  all  suffixes  from 
m  +  1  to  n.    The  equation  has,  however,  already  been  supposed 
true  for  suffixes  1  to  m  [cf.  equations  (170)  •  •  •,  (171)]. 
Thus  we  have  the  complete  system  of  equations 

d  (dL\      cL      ^  , 

dt\ddj      c^i 


d  /dL\      dL      ^  , 

it{^)-w,+^"'+''"-+ ■■■  =  '• 

in  which  the   suffixes   range  from  1  to  n.     On  eliminating  tlie 

m  multipHers  X,  /jl,  ■  •  ■  from  these  n  equations,  we  are  left  with 

n  —  m  equations,  wliich  enable  us  to  determine  tlie  changes  in 
the  coordinates. 


336 


GENERALIZED   COOEDIXATES 


ILLUSTRATIVE  EXAMPLES 

1.  A  homogeneoxis  sphere  of  radius  a  rolls  down  the  outer  surface  of  a  fixed 
sphere  of  radius  b  without  sliding.    Find  the  motion. 

At,  any  instant  let  the  line  of  centers  make  an  angle  0  with  the  vertical, 
and  let  the  angular  velocity  of  the  rolling  sphere  be  d.  The  velocity  of  the 
center  of  the  rolling  sphere  is  {a  +  b)  <p,  so  that 

T=  \m[{a  +  by-,f>^  +  ^a^e^]. 

The  potential  energy  is 

W  =  mg  {a  +  b)  cos  <f>, 
so  that  L=T-W 

=  ^m{a  +  i)202  +  1  ma^d^ 
—  mg  {a  +  b)  cos  </>.     (a) 

The  variations  in  0  and  0  are  not 
capable  of  having  any  values  we 
please,  for  the  velocity  of  the  center 
of  the  moving  sphere  Ls  (a  +  6)  0, 
and  also  mast  be  ad,  since  the  sphere 
is  rolling  vs^ith  angular  velocity  0 
without  sliding.    Thus 


Fig.  154 


ad  =  {a  +  b)(p. 


(b) 


This  is  true  at  every  instant  of  every  possible  motion,  so  that  we  must  have, 
on  integrating  with  respect  to  the  time, 

a9  =  {a  +  b)(f>  +  a  constant, 

■and  hence  we  must  suppose  changes  in  the  coordinates  0,  <f>  to  be  connected  by 

the  relation  ..      ,      ,  r^ » , 

aS0  =  (a  +  b)d<p. 

Thus  Lagrange's  equations  are 

d  /dL\  _  cL 


dt 


\d0/ 


C0 


+  Xa  =  0, 


d(dL\      cL      ^^        .^ 
dt  \c(f>/       c(j> 


0. 


Eliminating  X,  we  obtain 

^  ldt\d0/      B0\         ldt\d4>)      dcp] 

Substituting  from  equation  (a),  this  becomes 


(a  +  ^)  I  ^  (  J  »ia2  0 


d  12 


)]-[ 


{m{a  +  bY<p)  -  mg(a  +  b) 


sin  0=0. 


ILLUSTRATIVE  EXA:\IPLES 


After  replacing  ad  by  {a  +  h)  (p  from  equation  (?>),  we  liave 

-  ma  (a  +  b)^ =  mga  {a  +  b)  sin  0, 

6  dt- 


(a-J-6) 


;gsm<f>, 


showing  that  the  center  of  the  moving  sphere  moves  vpitli  five  seventlis  of  tlie 
acceleration  of  a  smootli  particle  sliding  down  a  sphere  of  radius  a  -r  b. 

The  same  result  could  have  been  obtained  by  eliminating  0  from  equations 
(a)  and  (6),  and  then  regarding  0  as  a  single  Lagrangian  coordinate. 

2.  A  flyioheel  is  connected  by  a  crank  and  rod  to  a,  piston  moving  in  a  hori- 
zontal cylinder.  When  there  is  no  steam  in  the  engine  the  flywheel  rests  in  its 
position  of  equilibrium.    Find  its  motion  if  displaced. 

Let  a,  b  denote  the  length  of  the  crank  and  rod,  and  let  6,  (p  be  the  angles 
they  make  with  the  horizontal  in  any  position  of  the  flywheel.  Then  the  posi- 
tion of  the  engine  is  known  fully  when  d  and  <p  are  known.    Not  only  do  the 


Fig.  155 


values  of  6  and  0  suffice  to  determine  the  position  of  the  engine,  but  if  we 
assign  arbitraiy  values  to  6  and  0,  we  do  not  necessarily  obtain  a  possible 
position  for  the  engine. 

The  velocity  of  rotation  of  flywheel,  axle,  and  crank  is  6,  so  that  the  kinetic 
energy  of  this  motion  is  ^  Id"^,  where  I  is  the  moment  of  inertia  of  this  i>art  of 
the  engine  about  the  axis  of  the  flywheel.  The  coordinates  of  the  center  of 
gravity  of  the  rod,  which  we  shall  assume  to  be  its  middle  point,  measured 
from  the  axis  of  the  flywheel,  are  : 

horizontal :    a  cos  0  +  lb  cos  0, 
vertical :         ,V  b  sin  0. 

Thus  the  velocity  of  its  center  of  gravity  has  components 
—  {asind  ■  e  +  i  6 sin 0  •  0) 


338  GENERALIZED   COOKDINATES 

horizontally,  and  a  6  cos  0  ■  <}>  vertically.    The  whole  velocity  v  of  the  center  of 
gravity  of  the  rod  is  therefore  given  by 

1)2  =  (a  sin  e  -d  +  \h  sin  0  •  0)2  +  ( .^  5  cos  0  •  0)2 
=  a2  sin2  e  ■  e-  +  ah  sin  6  sin  (p  ■  (pd  -\-  \  h-^'^. 

The  angular  velocity  of  the  rod  is  0,  and  its  radius  of  gyration  k  is  given  by 

3  \2/       12 
Thus,  if  m  is  the  mass  of  the  rod,  the  kinetic  energy  of  the  rod  is 

\  m  (u2  -f  fc202) 

=  I  m  (a2  sin2  6  6'^  +  ah  sin  ^  sin  0  00  +  i  ft2^2). 

Lastly,  the  horizontal  distance  of  the  end  of  the  piston  rod  from  the  center 

of  the  flywheel  is  a  cos  0  +  6  cos  0,  so  that  the  velocity  of  the  piston  and  piston 

rod  is 

—  a  sin  0  ■  6  —  b  sin  0  •  0. 

If  M  is  the  mass  of  the  piston  and  piston  rod,  the  kinetic  energy  of  this  part 

of  the  engine  is  ,,,,•„    ^      ,    •  \o 

^  ^l/(asin0- 0  +  6sin0- 0)2. 

We  now  have,  for  the  whole  kinetic  energy  T, 

2T=  Id-  +  m  (a2  sin2  d  ■  ff^  +  ab  sin  6  sin  0  ■  00  +  \  b^<p^) 

+  3/(asin0-0  +  6sin0- 0)2.  (a) 

The  potential  energy  TF,  measured  from  a  standard  configuration  in  which 
0  =  0  =  0,  IS  w  =  -  m:gh  sin  (0  +  e)  -  \  mgb  sin  0.  (6) 

Here  ^Mi  is  the  total  mass  of  flywheel  and  crank,  and  h,  e  are  the  polar  coor- 
dinates of  its  center  of  gravity  when  0  =  0. 

The  changes  in  6  and  0  are  not  independent.    A  glance  at  the  figure  shows 
that  we  must  always  have 

a  sin  0  =  6  sin  0,  (c) 

and  on  differentiating  this,  we  can  see  that  if  0  and  0  are  taken  as  generalized 
coordinates,  we  must  suppose  them  connected  by 

a  cos  0  30  —  6  cos  0  50  =  0. 
Thus  Lagrange's  equations  will  be 

d  /dL\      cL      ^  „      .  , ,, 

—  — T t-Xacos0=:O,  (d) 

c^A^0/     c0 


(    T )  - X6  cos  0  =  0.  (e) 


d/dL\_dL 


LAGEANGE'S  EQUATIONS  339 

The  elimination  of  \  from  these  equations  gives 

rd/cL\      cLl  Sd/cL\      cLl      ^ 

6  cos  <A    —  — r  I \  +  acoaei  —[^] =0, 

ldt\c0/       de]  ldt\c^/      c<p_\ 

and  on  substituting  for  L  from  equations  (a)  and  (b) ,  this  equation  becomes  an 
equation  between  6,  0,  and  their  differential  coeflBcients  with  respect  to  the  time. 
From  this  and  the  geometrical  relation  (c), 

a  sin  6  =  b  sin  (/>,  (/) 

we  can  proceed  to  determine  6  and  <p  in  terms  of  the  time. 
Using  equation  (/),  we  can  transform  equation  (6)  into 

W  =  —  ^gh  sin  (^  +  e)  —  -J  mga  sin  0. 

It  will  be  possible  to  arrange  countei-poises  on  the  flywheel  in  such  a  way  as 

to  make  ^      t-^,    ,   ,  „ 

e  =  0,     CMh  +  I  ma  =  0, 

and  if  this  is  done,  the  center  of  gravity  will  always  be  at  the  same  height. 
This  is  called  balancing  the  engine. 

If  we  suppose  the  engine  balanced  in  this  way,  we  have  W  =  0  and  there- 
fore L  =  T.  We  can,  however,  determine  the  motion  much  more  simply  than 
by  using  Lagrange's  equations,  for  we  know  that  T  must  remain  constant 
throughout  the  motion ;  and  by  differentiation  of  equation  (/)  we  have 

a  cos dd  —  b  cos <t)  <j>, 

so  that  we  can  replace  equation  (a)  by 

2T—  le^  +  m  (a2  sin2  e  ff-  +  a-  sin  d  cos  d  tan  0  6-  +  \  cC^  cos^  e  sec^  (p  ff^) 
+  M(a  sin  6*  •  ^  +  a  cos  ^  tan0  •  ey 
=  ^-[J  +  mar  sin  6  sin  (d  +  (j>)  sec  <p  +  \  ma^  cos^  0  sec^  0 
+  Ma^  sin2  {$  +  0)  sec^  0]. 

This  is  constant  throughout  the  motion,  but  we  see  that  it  does  not  follow 
that  6  is  constant.  Thus,  although  the  engine  is  balanced  so  as  to  remain  at  rest 
in  any  position,  it  will  not  necessarily  run  evenly  if  started  into  motion. 

Lagrange  s  Eqttations  for  Non-Conscrvativc  Systems 

273.  For  non-conservative  systems  it  has  been  shown  (§  268) 
that  equation  (In 6),  namely 


£ 


8Ldf  =  0,  (172) 

must  be  replaced  by 

r  \8T+^{X8,r+Y8i/  +  Z8z)]dt  =  0.  (173) 


340  GENERALIZED   COORDINATES 

Now  since,  as  iu  equation  (158), 

we  must  have 

Sx  =  x'  —  X 

neglecting  small  quantities  of  the  second  order. 

Thus 
^{Xhx  +  Yhy  +Zhz)  =  0,  W^+  0^  g6',+  .  •  .  +  e„86'„, 

where  0^,  ©2?  "  "  "^  ®»  depend  on  the  configuration  of  the  system, 
and  therefore  are  functions  oi  6^,  6^,  •  •  • ,  6^  only. 
Equation  (173)  now  becomes 

ChTdt  +  r  \®^he^  +  ^„je.,  +  •  •  •  +  ^JOjdt  =  0.  (174) 


Just  as  in  §  270  we  found  that 


£ 


hLdt 


could  be  transformed  into 


m 


he. 


c_L_d  IdL 
1^       'dt\^J 


dt, 


so  the  first  term  of  equation  (174)  can  now  be  transformed  into 


r{?'''K~^©j} 


dt. 


Substituting  this,  the  equation  becomes 
'^T  _d^/dT\ 


np'^ 


K    dt  \ddj 


+  01 


dt=  0. 


lagra:nge's  equations  341 

Since  this  is  true  for  all  possible  ranges  of  time,  we  must  have 


=  0  (175) 


at  every  instant. 

If  the  ^'s  can  vary  mdependently,  each  coefficient  must  vanish, 
and  the  system  of  equations  will  be 

|/^\      ^_@^  =  0,  (176) 

etc.,  while  if  W^,  80^,  •  •  •  are  connected  by  the  constraints  implied 
in  equations  (166),  (167),  •  •  •,  we  find,  as  in  §  272,  that  the  sys- 
tem of  equations  must  be  replaced  by 

d /dT\      dT      ^       ^  ,  ^  ,,^^^ 

dt  \ddj      c^x 

274.  These  systems  of  equations  reduce  to  tliose  previously 
obtamed  in  the  special  case  in  which  the  forces  are  conservative. 
For  in  this  case  consider  the  work  done  in  a  slight  displacement  hi 
which  9^  alone  varies,  6^  being  increased  by  hd^.     It  is  ^^86^,  and 

IS  also  —  —J-  oa    so  that  (s)^  =  —  — r-  • 

dT      ^        dT      dW      dL 

dT      d(L+W)       dL 
and  —  =  — !^ =  — ) 

dd^  dd^  dd^ 

since  W  does  not  contain  6^.    Thus  equation  (175)  reduces  to 

ii  I^J±\ _  ^  -  0 
dt  \^J      W,  ~    ' 

as  before,  and  m  the  same  way,  equations  (177),  •  •  •  can  be  trans- 
formed into  the  equations  obtained  in  §  272. 


342 


GENERALIZED  COORDINATES 


Lagrange's  Equations  hy  Direct  Transformation 

275.  Instead  of  deducing  Lagrange's  equations  from  equation 
(156),  they  may  be  obtained  directly  by  transformation  of  the 
equations  of  motion. 

We  have,  as  before, 

so  that,  on  differentiation, 


or 


dx  _  df_dd^       d^cW^ 
dt  ~  dd^  dt       dd^  dt 

dx  A        dx  A 


(178) 


(179) 


Thus  i;  is  a  linear  function  oi  6^,  B„,  ■  •  ■,  and 

dx  _  ox 

We  have  T  =  l  ^m  (i'  +  tf  +  z"), 

so  that  T,  as  before,  is  a  quadratic  function  of  6^,  6^,  -  -  -,  which 
also  involves  6^,  6^,  ■  ■  ■.    By  differentiation, 


oT 


=X 


dx        .  oil 
m  ( ;c  -7-  +  2/  -T-  + 


dz 


or,  by  equation  (179), 
cT 


30, 


dsQ        .    dy       .    dz 


Thus 


d  IdT 
dt 


(/^  \  _  ^     Id^x  dx       d^u  dy       d?z  dz 


+  S 


m 


df  dd^ 


dt  [ddj  ^-^  dt  VddJ  ^    dt  \dd^ 


(180) 


LAGRANGE'S  EQUATIONS  343 

Since  -jr-  is  a  function  of  6^,  0^,  •  •  •,  we  have 

dt\deJ~del~dF    ddje.'dt        '  ^     ^ 

while,  by  differentiation  of  equation  (178), 

The  right-hand  members  of  equations  (181)  and  (182)  are  seen 
to  be  identical,  so  that 

d  / dx\  _  dx 

and  the  last  line  of  equation  (180)  transforms  into 
.,r-v     / .  dx    ,    .    By    ^    .   dz\ 


•^m  [|  (;?  +  f  +  ?)], 


of  which  the  value  is 

dT 

Equation  (180)  now  becomes 

dt  \d^J      a^i  ~^'''  [df  dO,      df  86^      dt'  36, 

From  the  equations  of  motion, 

dh:     , 
X  =  m  —-Z )  etc., 
df 


so  tliat  tliis  again  becomes 

d 
dt 


d(dT\_dT^^/d^dy^dz\ 
it\ddj      dd,      ^\    cBj     cdj     cdj. 


344 


GENERALIZED   COORDINATES 


If  we  give  the  system  a  small  displacement,  in  which  6^  is 
increased  to  6^  +  86^,  6^  to  6^  +  80^,  etc.,  we  have,  on  equatmg 
two  different  expressions  for  the  work  done : 

S,8d,  +  ®,Bd,  +  .  • . 

=  ^{X8x  +  YSi/  +  Z8z). 

On  substituting  the  value  of  8x  obtained  on  page  340,  this 


=x 


x(|a..+  ^8..+ 


=  2^iX 


A' 


cd^ 


00.  d0. 


=  80, 


d  /dT\  _  dT 


+  5(9., 


_dt\ddj       d0^ 
and  the  equation  reduces  to 

dt  \dej    30. 


+ 


X80, 


-a 


0. 


This  is  the  same  equation  as  equation  (175),  and  the  different 
forms  of  Lagrange's  equations  can  be  deduced  as  before. 


Lagrange's  Equations  for  Impulsive  Forces 

276.  Let  a  system  of  impulses  act  during  the  short  interval 
from  t  =  t^  to  t  =  t^.  Let  0^,  0^,  ■  •  ■ ,  0^  now  be  supposed  to  be 
independent  coordinates,  so  that  Lagrange's  equations  are 

d  /dT\      dT       „       ^ 
-r\—^]  —  ^^  =  ®i>  etc. 
dt  \30j      d0. 

If  we  multiply  by  dt,  and  integrate  from  t  =  t^  to  t  =  t^,  we  have 


rm'-riH:"* 


The  value  of  the  first  term  is 

^dT\    .  _ 
\d0j  t=t2 


oO^/  t=ti 


LAGRANGE'S  EQUATIONS  345 

and  when  the  interval  from  t^  to  t^  is  made  vanishingly  small,  this 

dT 
measures  simply  the  change  in  —  produced  by  the  impulse. 

Jr'-  cT  dT 

—  (It,  the  integrand  ^^  is  finite,  so  that 


dd. 


when  the  interval  of  time  is  supposed  to  vanish,  this  term  will 
vanish  with  it.    Thus  the  equation  becomes 

change  m  ^^  =  j     S^dt  (183) 

277.  If  F  is  an  ordinary  force  acting  impulsively  through  the 
interval  t^  to  t^,  we  call  |    Fdi  the  impulse.    By  analogy  we  call 

2 

0,  dt 


•'2 


X 


the  generalized  impulse,  corresponding  to  the  generalized  coordi- 
nate 01.    Thus  we  have  equation  (183)  in  the  form 

d  T 
change  in  — -  =  generalized  impulse. 

From  analogy  with  the  relation, 

change  in  momentum  of  a  particle  =  impulse  on  particle, 

dT 
we  call  — r-  the  generalized  momentum  corresponding  to  the  coor- 

dinate    6^.     Thus    with    these    meanings  attached   to  the  terms 
"impulse"  and  "momentum,"  the  relation 

change  of  momentum  =  impulse 

is  true  in  generalized  coordinates. 

When  our  coordinates  are  x,  y,  z,  the  coordinates  in  space  of  a  moving 
particle,  the  generalized  nionient.i  will  of  course  become  identical  with  the 
ordinary  components  of  monieiituiu.    We  have 

dT 
so  that  —  =  mx,  etc. 

ex 


346 


GENEEALIZED  COORDINATES 


Euler's  Equations  for  a  Rigid  Body 

278.  Euler's  equations  (§  252)  can  be  derived  from  those  of 
Lagrange. 

Let  the  moments  of  a  rigid  body  about  its  principal  axes  of 
inertia  at  a  point  0,  which  is  fixed  in  the  body  and  is  also 
either  fixed  in  space  or  is  the  center  of  gravity  of  the  body,  be 
A,  B,  C.  Then  if  w^,  co^,  co^  are  the  components  of  rotation  about 
these  axes,  we  have,  as  in  §  248, 

T  =  1  {Ao^i  +  Bcol  +  Cwl ).  (184) 

As  Lagrangian  coordmates,  let  us  take  6,  (f)  the  spherical  polar 
coordinates  of  the  third  axis  OC  of  the  body,  and  i/r  a  third  coor- 
dinate which  measures  the  angle  between  the  first  axis  OA  of  the 

rigid  body  and  the  plane 
through  OC  and  the  axis 
^=0,  say  the  plane  COz. 
We  have  first  to  find 
a)j,  ft).,  and  (o^  in  terms  of 
6,  cf)  and  yjr,  so  as  to  ex- 
press 2T  as  a  function  of 
these  coordinates.  The 
motion  of  the  body  is  com- 
pounded of  the  motion 
relative  to  the  plane  COz, 
together  with  the  motion 
of  the  plane  COz  relative 
to  fixed  axes.  The  former 
motion  consists  of  a  rotation  t/t  about  OC,  and  this,  resolved  along 
the  axes  OA,  OB,  OC,  has  components 

0,  0,         f. 

Tlie  motion  of  the  plane  COz  is  compounded  of 

{a)  a  rotation  6  about  an  axis  at  right  angles  to  its  plane ; 
(b)  a  rotation  (j)  about  the  axis  ^  =  0. 


Fig.  156 


EULER'S  EQUATIONS  FOR  A  RIGID  BODY       347 

Eesolved  along  the  axes  OA,  OB,  OC,  the  first  part  has  compo- 
nents 

0  sin  -yjr,  0  COS  -yjr,  0, 

while  the  second  has  components 

—  (f>  sin  0  cos  yjr,  (p  sin  6  sin  yjr,         <f>  cos  6. 

Compounding  these  motions,  we  obtain 

(o^  =  6  sin  y(r  —  ^  sin  6  cos  yjr 


(185) 


(o„  =  6  cos  yjr  -j-  (j)  sin  0  sin  -^ 
tWg  =  i/r  +  (^  cos  ^ 

Let  the  work  done  in  a  small  displacement  be 
@80  +<P84,  +  '¥8f; 
then  Lagrange's  equation  for  the  coordinate  ^  is 

We  have,  by  differentiation  of  equation  (184), 

dT  d(o  d(o,  d(o^ 

—^  =  Aq)^  — 7^  +  B(0^  — r-  +  CtWj  — r- 

on  substituting  from  equations  (185).    Also 

—  =A(o^  — i  +  Bo).,  — ^  +  C&>3  — ^ 

=  Ao)^  (0  cos  t/t  +  ^  sin  ^  sin  i/r) 

+  ^ty.,  (—  6)  sin  i/r  +  (^  sin  ^  cos  i/r) 
=  (^  —  ^)  a)j(u.,. 

Finally  "^Si/r  is  the  work  done  by  external  forces  in  a  small 
rotation  8-v|r,  and  therefore,  by  §  121,  "^  is  equal  to  N,  the  sum  of 
the  moments  of  these  forces  about  the  axis  OC. 


348  GENERALIZED  COORDINATES 

Making  all  these  substitutions,  equation  (186)  becomes 
c'^-{A-B)<o,co,  =  N, 

wliicli  is  Euler's  tliird  equation,  and  the   other   two    equations 
follow  from  symmetry. 


Small  Oscillations 

279.  Let  ^1,  ^2'  ■  ■  ■'  ^«  ^^®  generalized  coordinates  of  any  system, 
and  let  it  be  supposed  that  these  coordinates  are  all  independent, 
so  that  any  set  of  values  of  d^,  6^,  ■  •  • ,  ^„  gives  a  possible  configura- 
tion of  the  system. 

Suppose  that  the  configuration 

d,^e^,       e,=^e,,       ■••,      e,  =  e,^  (is?) 

is  known  to  be  a  configuration  of  equilibrium.    Then,  if 

the  quantities  ^^,  <^2>  '  "  " »  ^n  ^^J  ^^^  taken  to  be  generalized  coordi- 
nates of  the  system,  and  will  possess  the  property  of  all  vanishing 
in  the  position  of  equilibrium. 

Let  Wf^  denote  the  value  of  the  potential  energy  in  the  configu- 
ration of  equilibrium.  The  potential  energy  in  any  other  configu- 
ration may,  by  Taylor's  theorem,  be  expanded  in  the  form 


where  all  the  differential  coefficients  are  evaluated  in  the  position 
of  equilibrium.  In  this  position  of  equilibrium,  however,  we  have, 
by  the  theorem  of  §  135, 

dW^  _dW__        _gTr  _ 


SMALL  OSCILLATIONS  349 

so  that  we  can  write  the  vahie  of  IF  in  the  form 

W=W,  +  a,,^\  +  2  a,,cl>,4>,  +  •  •  •  +  «„„(/>f„  (188) 

in  which  powers  of  ^^,  <j>2,  •  ■  ■  higher  than  the  second  are  left  out 
of  account,  because  we  are  going  to  confine  our  attention  to  motions 
in  which  (f)^,  (f)^,  -  ■  •  are  aU  small  quantities. 

The  kinetic  energy,  as  before  {§  270),  is  a  quadratic  function  of 
^v  <f>2'  •  ■  •>  i>n-    Let  us  say 

T  =  h,,cj>l  +  2  hj,4>,  +  •..  +  L„,^<f>l  (189) 

The  coefficients  Jj^,  h^^^,  •  ■  •  ,l>„„  are,  strictly  speaking,  functions  of 
^1)  ^2'  ■  *  ■'  ^»'  b^it  ^^6  ^^y  regard  their  values  as  bemg  equal  to 
the  values  in  the  configuration  of  equilibrium,  and  so  may  treat 
them  as  constants. 

280.  Now  consider  two  quadratic  functions  of  n  variables  x^, 
^2J  •  ■  ■  >  ^n»  defined  by 

F{x^,  x^,     ",  x^)  =  h^^^r^  +  2  h^^x^x^  -{ f-  &„„a;. 

Since  the  function  T  defined  by  equation  (189)  is  necessarily 
positive,  it  follows  that  F{x^,  x^,  •  •  •,  x^)  is  positive  for  all  values 
of  ajj,  x^,  •  •  -,  x^.  Hence,  by  a  known  theorem  in  algebra,  we  can. 
find  a  transformation  of  the  type 

1       iisi       i_b.  I  ^^^^^ 

in  which  the  coefficients  k^^,  etc.,  are  real,  which  is  such  that/  and 
F  transform  into  expressions  of  the  type 

f{x^,  x„,  ■■■,  xj  =  a,^\  +  a.^:,  +  ■■■  +  a,^^l, 

F{x,,  x„  .  .  .,  x„)  =  Afi  +  AI2+  •  •  •  +  /3n^l 

and  all  the  coefficients  ^^,  /S.,,  ■  •  • ,  ^„  will  be  positive. 


350  GENERALIZED  COORDINATES 

Algebraic  proofs  of  this  theorem  will  be  found  in  treatises  on  analysis, 
or  in  Salmon's  Hi(/her  Algebra,  Lesson  VI.  The  theorem  will  be  readily 
understood  on  considering  a  geometrical  interpretation  in  the  case  in  which 
the  number  of  variables  is  tliree.    Calling  the  variables  x,  y,  and  z,  the 

equations 

/(x,y,  2)  =  1,     F{x,y,z)^\  (191) 

will  be  the  equations  of  concentric  quadrics  ;  and  since  F  is  positive  for  all 
values  of  x,  y,  z,  the  second  quadric  will  be  an  ellipsoid.  It  is  known  that 
two  concentric  quadrics,  of  which  one  is  an  ellipsoid,  always  have  one  real 
set  of  mutually  conjugate  diameters  in  common.  A  transformation  of  the 
type  expressed  by  equation  (191)  enables  us  to  transform  to  these  axes  as 
axes  of  coordinates,  and  the  equations  of  the  quadrics  are  then  of  the 
required  forms 

a,e^  +  a,^l  +  a,^l  =  1 ,     p^^i  +  /3,,fi  +  ^^^l  =  1.  (192) 

[Simple  reasoning  will  show  the  truth  of  the  geometrical  theorem  that 
an  ellipsoid  and  a  second  quadric  always  have  one  common  set  of  real 
mutually  conjugate  diameters.  For  a  real  linear  transformation  will  trans- 
form the  ellipsoid  into  a  sphere,  and  the  second  quadric  into  a  new,  but 
still  real,  quadric.  The  jjrincipal  axes  of  this  real  quadric  are  now  real 
mutually  conjugate  diameters  for  the  sphere  and  the  quadric,  and  on  trans- 
forming back,  real  mutually  conjugate  diameters  remain  real  mutually 
conjugate  diameters.] 

The  algebraic  proof  that  equations  (191)  could  be  transformed  into 
equations  (192)  would,  however,  clearly  not  be  limited  to  the  case  of  three 
variables,  so  that  the  theorem  must  be  true  for  any  number  of  variables. 

281.  This  theorem  proves  that  we  can  find  new  coordinates 
yjr-^,  yjr.-,,  •  •  • ,  "v/^,,  connected  with  ^j,  (j)^,  •  •  •,  <^„  by  relations  of  the 
type 

<t>l  =  /^llfl  +  f^uf^  +   •   •   •   +  '<^l„fn,  (193) 

4  =  '^ll^l  +  «12t2  +  ---   +  «l»^n.  (194) 

such  that,  expressed  m  terms  of  these  coordinates,  the  potential 
and  kinetic  energies  assume  the  forms 

W=iV,+  a^fl  +  a^  +■■■  +  a^^lrl,  (195) 

T  =  ^,irl  +  /3^  +■■■+  ^^yjrl  (196) 

The  coordinates  yfr^,  i^„,  •  ■  • ,  "v/^,,  are  called  the  principal  coordi- 
nates of  the  system,  or,  by  some  writers,  the  norinal  coordinates. 


SMALL  OSCILLATIONS  351 

Lagrange's  equations,  in  terms  of  these  coordinates,  are 
^  l^I\      ^T  _      cW 

which  become  /Qi  — -V  =  —  «i'^p  etc.  (197) 

at 


Stable  Eqtiilibrmm 

(X 

282.  If  QTi  is  positive,  let  us  put  —i  =  A^,  so  that  k^  wiU  be 
real.    The  equation  is  now  ^ 

d'i^l  7  2    f 

of  which  the  solution  is 

ylr^  =  A^  cos  {k^t  —  e;},  (198) 

as  in  §  208.  Thus  the  motion  is  a  simple  harmonic  motion  of 
frequency  k^.  If  all  the  coefficients  «,,  «„,  •  •  •,  «„  are  positive,  the 
complete  solution  of  the  equations  will  be  of  the  form 

yjr^  =  A^  cos{k^t  —  e^), 

^|r„  =  A^  cos  {k^t  —  €^),  etC, 

and  the  coordinate  x  of  any  particle,  of  wliich  the  value  in  the 
equilibrium  position  is  x^,  w^iU  be 

^  =  ^-,  +  (^,-^;)^  +  (^,-^",)^^+... 

,     dx         ,     dx 

=  Xq  +  B^  cos {k^t  —  e J  +  i>2  cos (kj  —  e^)  +  •  •  •, 

where  B^,  B„,  ■  ■  ■  are  new  constants. 

Tlius  the  motion  of  any  single  particle  will  be  a  motion  com- 
pounded of  a  number  of  simple  harmonic  motions. 


352  GENERALIZED  COORDINATES 

283.  The  potential  energy  corresponding  to  any  principal  coor- 
dinate yjr^  is  a^-sjrl,  or,  if  we  suppose  y}r^  given  by  equation  (198),  is 

Similarly,  the  kinetic  energy  correspouding  to    tliis  principal 
vibration  is  ^.yjrt  or 

Averaged  over  a  very  long  time,  the  average  values  of  cos^  (h^t  —  e^) 
and  of  sin^(Z;^?;  —  Cj)  are  each  -|,  so  that  the  average  potential  and 
kinetic  energies  are  respectively 

(X 

and  these  are  equal  since  A?  =  -f  •  Thus  in  any  vibration  the  aver- 
age  kinetic  and  potential  energies  are  equal. 


Unstable  Equilibrium 

284.   Suppose  now  that  any  one  of  the  coefficients  in  equation 

cc 
(195)  is  negative,  say  a^.    Let  us  put  — i  =—  1^,  so  that  ^\  wiU  be 

Pi 
real.    Equation  (197)  now  assumes  the  form 

dt-   -^'^'' 
and  this  has  as  solution 

^|r^  =  A^e^''  +  B^e''"'*, 

showing  that  -yfr^  increases  indefinitely  with  the  time,  and  does  not 
oscillate  about  the  value  yjr^  =  0.  Thus  the  motion  is  unstable,  and 
we  now  see  that  the  motion  can  only  be  stable  provided  all  the 
coefficients  a^,  a„,  •  •  •,  a;„  are  positive.    In  other  words. 

For  stable  equilibrium  the  jJotential  energy  in  the  configuration 
of  equilibrium  must  be  an  absolute  minimum. 

This  is  the  result  which  has  already  been  stated  without  proof 
in  §  153. 


FOKCED  OSCILLATIONS  353 

Forced  Oscillations 

285.  The  oscillations  which  have  so  far  been  considered  are  of 
the  type  known  as  free  vibrations,  —  that  is  to  say,  the  forces  act- 
ing arise  entirely  from  the  potential  energy  of  the  system  itself. 

A  second  type  of  oscillation  occurs  when  the  system  is  acted  on 
by  forces  from  outside,  in  addition  to  those  arising  from  its  own 
potential  energy.    These  oscillations  are  known  as  forced  oscillations. 

Let  us  suppose  that  the  potential  and  kinetic  energies  of  tlie 
system  are  given  by  equations  (195)  and  (196),  and  that  tlie  sys- 
tem of  external  forces  acting  at  any  instant  is  such  that  the  work 
done  m  a  small  displacement  is 

^jSi/tj  +  ^p.^  gi^._, -I . 

Then  Lagrange's  equations  for  this  system  are 

dt  \dyjrj       ^if'i  ^"f  1  ^' 

which  becomes  2  /3,  — ^  =  —  2  a^-^^  +  ^,,  (199). 

in  which  "^j,  it  must  be  remembered,  is  now  a  function  of  the 
time.  This  equation  can  be  solved  according  to  the  rules  given 
in  any  treatise  on  differential  equations.    If,  as  before,  we  take 

cc 
kl  =  -^  >  the  general  solution  is  found  to  be 

Pi 

r  —  t 

ti  =  A,  cos  {k,t  -  e,)  +  -J=       f  i^,),^  ,sin  k,  (f  -  f)  dt', 

r  =  -  » 

the  lower  limit  of  integration  being  either  t'  =—  cc,  or  the  instant 
of  wliich  the  external  forces  first  came  into  operation. 

286.  A  case  of  extreme  importance  occurs  when  ^^  is  simply 
periodic  with  respect  to  the  time,  say 

■^^  =  -£■  cos  (2)^1  —  7i). 


354  (JENERALIZED  COORDINATES 

The  solution  is  then  found  to  Ije 


or,  since  a^  =  ^J^l, 

f^  =  Jj  COS(/'i^  -  €,)  + —  cos{p,t  -  7,). 


Thus  the  variation  in  -\/r^  is  now  compounded  of  a  simple 
harmonic  motion  of  frequency  k\,  and  also  one  of  frequency  2h, 
the  frequency  of  the  impressed  force. 

AVe  notice  that  if  p^  is  very  nearly  equal  to  l\,  then  the  second 
vibration  is  of  very  large  amplitude.  In  the  limiting  case  in  which 
Pi  =  A\,  the  amplitude  of  the  second  vibration  becomes  mfuiite, 
but  now  the  two  vibrations  are  of  the  same  period,  so  that  they 
may  be  compounded,  and  we  cannot  say  that  the  resultant  vibration 
is  one  of  infinite  amplitude,  because  we  do  not  know  the  values  of  J^ 
and  ej,  and  these  may  just  be  such  as  to  destroy  the  infinite  ampli- 
tude of  the  second  term.  The  result  we  have  obtained  may  be 
enmiciated  in  the  following  form : 

When  a  system  is  acted  on  hy  a  periodic  force,  of  frequency 
very  nearly  equal  to  that  of  one  of  the  principal  vibrations  of  the 
system,  then  the  forced  oscillations  vnll  he  of  very  great  amplitude. 

This  is  known  as  the  jJ^inciple  of  resonance. 

The  principle  is  one  of  which  many  applications  appear  in  nature.  For 
instance,  a  bridge,  not  being  absolutely  rigid,  may  be  regarded  as  a  system 
having  a  number  of  free  vibrations.  A  body  of  men  marching  over  the 
bridge  in  regular  step  will  apply  a  periodic  force,  and  if  the  period  of  their 
step  happens  to  nearly  coincide  with  one  of  the  free  periods  of  the  bridge, 
the  amplitude  of  the  vibrations  forced  in  the  bridge  may  be  so  large  as  to 
endanger  the  bridge.  For  this  reason  troops  are  ordered  to  "  break  step  " 
when  crossing  a  bridge. 

Again,  a  ship  is  not  perfectly  rigid,  and  so  will  possess  a  number  of  free 
vibrations.  The  motion  of  its  engines  will  apply  a  pei-iodic  force  of  period 
equal  to  that  of  its  revolution,  and  if  this  coincides  with  that  of  one  of  the 
vibrations  of  the  ship,  large  pulsations  will  be  set  up.  This  can  be  reme- 
died by  altering  the  speed  of  the  engine  until  it  no  longer  is  in  resonance 
with  the  free  \dbrations  of  the  ship. 


THE  CANONICAL  EQUATIONS  355 

As  a  last  example,  it  may  be  noticed  that  a  sliip  will  haA'e  a  free  period 
of  rolling  about  its  vertical  jiosition.  If  it  is  in  a  rolling  sea,  the  waves 
which  meet  it  will  apply  external  forces  which  may  be  regarded  as  approxi- 
mately periodic.  If  the  period  of  the  waves  happens  to  coincide  with  that 
of  the  ship,  the  ship  will  roll  heavily  even  though  the  waves  may  be  com- 
paratively small.  This  danger  can  be  remedied  by  altering  the  course  of 
the  ship  and  so  cavising  it  to  meet  the  waves  at  a  different  interval.  Another 
way  is  to  give  the  ship  a  list  by  spreading  canvas,  and  so  causing  it  to 
oscillate  about  a  different  position  of  equilibrium,  about  which  the  periods 
of  free  vibrations  are  different. 


The  Canonical  Equations 

287.  li  6^,  d„,  ■  ■  ■  are  Lagrangian  coordinates  of  any  system,  the 
kinetic  energy  T  is  a  quadratic  function  oi  6^,  6^,  6^,  •  •  ■.  Let  the 
corresponding  momenta  be  i(-^,  ii^,  ■  ■  ■,  «,,,  these  being  given  by 

dT 


u.  = 


ct 


etc.  (200) 


Now  let  us  introduce  a  function  T' ,  defined  by 

T'  =  u,e^  +  uA  + T, 

so  that  T'  is  a  function  of  i'^,  v.-,,  ■  ■  ■,  6^,  6.,,  ■  ■  ■,  6^,  0„,  ■  ■  • ;  and 
Wp  ^2,  •  •  •  are  of  course  functions  of  6^,  6.^,  ■  ■  ■,  6^,  6^,    ■  •. 

On  differentiation  of  T'  we  have 

dT'  =  1(^(16^  +  vjW.^  +  ■■■ 

+  6/h(^  +  d„dn^  -\ 

dT  dT 

and  this,  by  equation  (200),  reduces  to 

dT'  =  e^du,  +  0Ju.^  + ^  d0,  -^^dO.. .      (201) 

Cu.  Co... 


356  GENERALIZED  COORDINATES 

Since  the  differentials  cW^,  cW^,  •  •  •  do  not  occur,  it  appears  that  T^ 
can  be  expressed  as  a  function  of  it^,  ti^,  •  •  •,  d^,  6^,  ■  •  •  only.  We 
can  easily  find  its  value;  we  have 

=  2  T,  since  T  is  a  homogeneous  quadratic  func- 
tion of  e^,e^,---  +  6'„. 

Thus  T'  =  2  T  -  T  =  T, 

showing  that  T'  is  equal  to  T,  but  is  expressed  as  a  function  of 
U\}  ^2,  ■  •  -,  ^i,  ^2'  ■  ■  ■•    Thus 

To  illustrate,  let 

so  that  «i  =  2  (oil  +  hdo),     u^  =  2  (hdi  +  he..) . 

Then,  by  definition, 

T'  =  uiSi  +  W24  + T 

=  2ei  (a^i  +  he^)  +  2  ^2  (^"^1  +  ^^2)  -  ^ 
=  T=i  u^e^  +  i  u^e„. 

From  equation  (201),  we  have 

cT'  _  _dT 

dT' 

-^  =  e,.  (202) 

In  Lagrange's  equations 

dt\ddj      cd^~ 

dL       8(T-W)       dT 

we  have  — r-  =  -^ : =  -^  =  u.^- 

36,  cd,  cd. 


THE  CANONICAL  EQUATIONS  357 

Thus  Lagrange's  equations  may  be  written  as 

du^_cL  _d{T-W)  ^      d(T'+W) 
dt  "  cO^  ~        cd^       ~  dd^ 

while,  by  equation  (202),  —-^  = 

If  we  write  H  =  T'  +  W,  these  equations  assume  the  symmet- 
rical form 


dO,  _  du 

dK 

i^ 

dt       ou^ 

At 

l>^v 

du,           oH 

-^  =  —  ^TTT  '  etc. 
dt            cd^ 

J  t^... 

288.  Tills  is  known  as  the  canonical  form  of  the  dynamical 
equations.  The  function  H  is  called  the  Hamiltonian  function,  and 
since  H  =  T'  +  W,  we  notice  that  H  is  the  total  energy  expressed 
as  a  fimction  of  the  coordinates  d^,6^,-  ■  ■  ,6^^  and  of  the  momenta 

The  canonical  form  is  the  simplest  and  most  perfect  form  in 
which  the  generalized  dynamical  equations  can  be  expressed.  For 
this  reason  the  canonical  system  of  equations  forms  the  starting 
point  of  a  great  many  investigations  in  higher  dynamics,  mathe- 
matical physics,  and  mathematical  astronomy. 

289.  We  may  appropriately  terminate  the  present  book  by  giv- 
ing illustrations  of  the  use  of  generalized  coordinates  from  two 
branches  of  mathematical  physics. 

Illustration  from  hydrodynamics.  Let  a  solid  of  any  sliape  be  in  a  stream  of 
water  flowing  witli  uniform  velocity  V.  If  the  solid  is  at  a  sufficient  depth  from 
the  surface,  its  presence  will  not  disturb  the  flow  at  the  surface,  and  the  only 
disturbance  in  the  flow  of  the  water  will  be  in  the  neighborhood  of  the  soliil. 
It  can  be  proved,  from  elementary  hydrodynamical  principles,  that  there  is 
only  one  way  in  which  the  water  can  flow  past  the  solid.  Hence  it  follows 
that  the  kinetic  energy  of  the  flow  of  the  w'ater  is  given  by 


358  GENERALIZED  COORDINATES 

where  To  is  the  vahie  which  the  kinetic  energy  would  have  if  the  solid  were 
removed.  Suppose  that  the  solid  is  acted  on  by  external  forces,  besides  the  pres- 
sure of  the  water.  Let  the  sum  of  the  moments  of  these  forces  about  any  axis 
be  0,  and  let  ^  be  a  coordinate  which  measures  the  angle  turned  through  about 
this  axis.    Then  Lagrange's  equation  corresponding  to  the  coordinate  6  is 

dt\8e/       dd 
If  the  external  forces  just  suffice  to  hold  the  body  at  rest  in  the  liquid,  we 
have  —  ( — r  )  =  0,  so  that 

dt\8e)  ar       aa^, 

0  = = F2. 

cd  80 


Hence  the  sum  of  the  moments  of  the  liquid  pressure  must  be  —  0,  or 

dd 

We  can  calculate  a  from  the  shape  of  the  solid,  and  .so  can  obtain  a  knowledge 
of  the  couples  acting  on  the  solid. 

Illustration  from  electromagnetism.  The  energy  required  to  establish  the  flow 
of  two  steady  currents  of  electricity  of  strengths  i,  V  in  two  given  closed  circuits 
is  known  to  be  of  the  form 

E=  i  Li^  +  Mil'  +  -1-  Ni"^, 

where  L  and  N  depend  on  the  .shape  of  the  first  and  second  circuits  respectively, 
while  M  depends  on  the  shape  of  both  circuits,  and  also  on  their  positions 
relative  to  one  another. 

Suppose  that  the  second  circuit  is  free  to  move  along  any  line  towards  the 
first  circuit.  Let  x  be  a  coordinate  measured  along  this  line,  and  let  the  force 
required  to  hold  the  second  circuit  at  rest  be  X  in  the  direction  in  which  x  is 
mea.sured. 

Let  L  denote  the  usual  function  T  —  TF,  and  let  the  second  circuit  be  acted 
on  by  an  externally  applied  force  X.  Then  Lagrange's  equation  for  the  coor- 
dinate X  is 

d  /dL\  _£L_ 

dt\dx/       dx      "  ' 
so  that,  since  there  is  no  acceleration, 

X  =  _^. 

dx 

As  a  matter  of  experiment,  it  is  found  that 

^          ..,dM         dE 
X  =  —  II  —  = 

dx  dx 


EXAMPLES  309 

If  the  energy  of  the  two  currents  were  potential  energy,  we  should  have 
_c_L_      dW      _^__cW 
ex  ex  ex  ex 

so  that  the  force  X  would  be  exactly  opposite  to  that  observed. 
On  the  other  hand,  if  the  energy  is  kinetic  energy,  we  have 

_cL  __eT  __eE 
ex  ex  ex 

so  that  the  value  of  X  agrees  with  that  observed. 

Hence  we  conclude  that  the  energy  of  an  electric  cun-ent  is  wholly  kinetic. 

GENERAL  EXAMPLES 

1.  The  friction  of  an  engine  is  such  that  one  horse  power  can  run  it  at 
250  revolutions  per  second  when  it  is  doing  no  external  work.  The  inertia 
of  its  moving  parts  is  such  that  when  running  at  125  revolutions  per  sec- 
ond, and  acted  on  by  one  horse  power,  its  speed  is  accelerated  at  the  rate 
of  10  revolutions  per  second.  If  the  engine  is  left  to  itself  when  running 
at  its  full  speed  of  250  revolutions  per  second,  find  liow  many  revolutions 
it  will  make  before  coming  to  rest. 

2.  A  square  is  moving  freely  about  a  diagonal  with  angular  velocity  w, 
when  suddenly  one  of  the  angular  points  not  in  that  diagonal  becomes  fixed. 
Determine  the  impulsive  pressui'e  on  the  fixed  point,  and  show  that  the 
new  angular  velocity  will  be  1  w. 

3.  Four  equal  rods,  each  of  length  2  a  and  mass  yi,  are  freely  jointed  so 
as  to  form  a  rhombus.  The  system  falls  from  rest  wdth  one  diagonal  \er- 
tical,  and  strikes  a  fixed  horizontal  inelastic  plane.  Find  the  impulse  and 
the  subsequent  motion. 

4.  Two  particles  connected  by  a  rigid  rod  move  on  a  smooth  vertical 
circle.    Find  tlie  time  of  a  small  oscillation. 

5.  A  uniform  rod  of  length  /  has  the  two  points  at  distance  c  iroui 
its  middle  point  connected  by  equal  strings  of  length  L  to  two  fixed  ]>niuts 
at  distances  2  c  apart  in  the  same  horizontal  line. 

Find  the  principal  coordinates  and  the  corresponding  periods  of 
vibration. 

6.  If  the  rod  of  the  last  question  receives  a  horizontal  blow  of  iin]»ulse 
/  at  one  extremity  and  at  right  angles  to  its  length,  find  the  subsequent 
motion. 

7.  A  rough  uniform  cylinder  of  radius  a  has  an  inextensible  string  coiled 
round  its  central  section.  One  end  of  the  string  is  fastened  to  a  fixed  point 
P,  and  the  cylinder  is  rolled  up  the  string  until  it  is  touching  P,  with  tin- 
tangent  to  the  cylinder  at  P  vertical.  The  cylinder  is  then  let  go.  Find  the 
motion. 


360  GENERALIZED  COORDINATES 

8.  In  the  last  question,  find  the  motion  if  the  tangent  at  P  is  perpen- 
dicular to  the  axis  of  the  cylinder,  but  is  not  quite  vertical. 

9.  In  spherical  polar  coordinates  prove  that  the  kinetic  energy  of  a 
moving  pai^ticle  of  unit  mass  is  given  by 

T=   |(r2  +  7-2^2  +    ?-2sin2^02). 

Hence,  prove  that  the  acceleration  of  the  particle  has  components  in  the 
direction  of  /•,  6,  </>  increasing,  of  amounts 


(I_  /cT\  _  fT         1 
dt  \  cr  I       dr  r 


'1  /£L\  -  ^Tl  1        fl  (cT\ 

Jty^e)     cey      r  sin e  Jt  \J^) ' 


lit 
Show  that  the  actual  values  of  these  accelerations  are 

dfi  ^        r  dr      ■'  rsinedr  ^^ 

10.  The  velocity  of  a  particle  in  its  orbit  is  found  to  vary  in  the  inverse 
square  of  its  distance  from  a  fixed  point.  Ajiply  the  principle  of  least 
action  to  find  the  orbit,  and  thence  the  law  of  attraction. 

Deduce  the  same  results  from  the  law  of  conservation  of  energy. 

11.  Suppose  that  all  forces  are  annihilated  in  the  universe,  and  that 
there  is  a  concealed  mechanism  capable  of  possessing  kinetic  energy.  Sup- 
pose that  the  amount  of  this  kinetic  energy  depends  only  on  the  positions 
of  the  material  bodies  in  the  universe,  being  equal  in  magnitude  except  for 
a  constant,  and  opposite  in  sign,  to  the  potential  energy  which  the  system 
would  have  if  the  forces  had  not  been  annihilated. 

Show  that  the  dynamical  phenomena  of  a  universe  of  this  kind  will  be 
identical  with  those  of  a  universe  in  which  both  forces  and  kinetic  energy 
exist,  the  changes  in  the  latter  being  determined  by  Newton's  laws  of 
motion. 

12.  A  number  of  spheres  without  mass,  of  radii  a,  b,  c,  •  •  •,  move  in  a 
straight  line  through  an  infinite  ocean  of  density  p^ ,  the  distances  apart  of 
their  centers  being  ?-^ ,  r^^,  etc.,  and  their  velocities  ??„,  v^,,  v^,  ■  ■  ■ .  When 
a,  b,  c,  ■  •  ■  are  small  compared  to  r^,  etc.,  the  kinetic  energy  of  the  motion 
of  the  ocean  is  given  by 

Show  that  to  an  observer  who  is  unconscious  of  the  presence  of  the  ocean, 
the  spheres  will  appear  to  move  as  though  having  masses  |  Trpa^ ,  |  wpb^ , 
etc.,  and  as  though  forces  of  attraction  acted  between  every  pair  of  spheres, 
proportional  to  the  product  of  the  masses,  to  the  product  of  their  veloci- 
ties, and  to  the  inverse  fourth  power  of  the  distance  between  them. 


INDEX 


(The  munbers  refer  to  pages.) 


Absolute  units,  of  force,  30  ;  of  work, 
UV>. 

Acceleration,  12  ;  parallelogram  of,  13 ; 
in  circular  motion,  14,  18. 

Action,  328 ;  principle  of  least,  328. 

Amplitude,  of  a  pendulum,  261 ;  of  sim- 
ple harmonic  motion,  265. 

Angle  of  friction,  47. 

Angular  momentum,  297;  conservation 
of,  207. 

Angular  velocity,  286 ;  composition  of, 
287. 

Arc,  center  of  gravitj^  of  circular,  125. 

Atwood's  machine,  195. 

Average  velocity,  6. 

Axes  of  inertia,  303. 

Axis  of  rotation,  92. 

Balancing,  of  an  engine,  337. 

Belt,  center  of  gravity  of  spherical,  130. 

Canonical  equations,  355. 

Catenary,  80. 

Center  of  force,  motion  of  point  about, 
269. 

Center  of  gravity,  117;  of  a  lamina, 
121,  135;  of  a  solid,  132,  135;  of  a 
triangle,  121 ;  of  a  pyramid,  132  ;  of 
a  circular  arc,  126  ;  of  a  segment  and 
sector  of  a  circle,  128,  129;  of  a 
spherical  belt  and  cap,  130 ;  of  a  sec- 
tor of  a  sphere,  130 ;  motion  of,  of  a 
system,  224. 

Central  axis,  of  a  system  of  forces,  107. 

Centroid,  20. 


Circular  arc,  center  of  gravity  of,  125. 

Coefficient,  of  friction,  47 ;  of  elas- 
ticity,  24lt. 

Coefficients  of  inertia,  301. 

Composition,  of  motions,  4 ;  of  veloci- 
ties, 7 ;  of  accelerations,  13  ;  of  forces 
acting  on  a  particle,  37 ;  of  forces 
acting  in  a  plane,  95;  of  parallel 
forces,  99 ;  of  couples,  105 ;  of  rota- 
tions, 286. 

Compression,  moment  of  greatest,  238. 

Conical  pendulum,  271. 

Conservation,  of  energy,  171 ;  of  linear 
momentum,  223 ;  of  angular  momen- 
tum, 297. 

Conservative  system,  of  forces,  103. 

Coordinates,  generalized,  320,  320;  nor- 
mal or  principal,  350. 

Couples,  101;  in  parallel  planes,  104; 
composition  of,  105 ;  work  performed 
against,  154. 

Cycloidal  pendulum,  265. 

Degrees  of  freedom,  number  of,  184, 

332. 
Descent,  line  of  quickest,  193. 
Diagram,  indicator,  151. 
Differential  equations,  of  orbits,  275. 
Double  stars.  280. 

Earth's  rotation,  198,  310. 
Elasticity,  of  a  string,  45 ;   modulus 

of,  45 ;   of  a  solid,  238 ;   coeflBcient 

of,  240. 
Ellipsoid  of  inertia.  302. 


361 


INDEX 


Energy,  potential,  103;  kinetic,  168, 
228  ;  total,  171;  conservation  of ,  171; 
of  motion  of  a  system,  230 ;  of  a  rigid 
body,  290. 

Envelope  of  paths,  of  i)rojectiles,  211. 

Equation,  of  energy,  171,  250;  of  mo- 
tion of  a  particle,  254 ;  of  orbit  of 
a  particle,  275 ;  of  a  rigid  body,  304. 

Equations,  Euler's,  306,  340  ;  La- 
grange's, 329,  342  ;  canonical  (Ham- 
ilton'.s),  355. 

Equilibrium,  of  a  particle,  38,  41 ;  of 
a  system  of  particles,  03 ;  of  a  rigid 
body,  93 ;  stability  and  instability 
of,  174,  351. 

Euler's  equations,  300,  346. 

Extensibility,  of  strings,  44. 

Flexibility,  of  strings,  43. 

Force,  20  ;  measurement  of,  30  ;  trans- 
missibility  of,  94. 

Forced  oscillations,  353. 

Forces,  composition  and  resolution  of, 
37-39  ;  in  one  plane,  66,  95  ;  parallel, 
96,  99 ;  in  space,  106 ;  imi)ulsive, 
233. 

Frame  of  reference,  3,  33 ;  motion  re- 
ferred to  moving,  197 ;  kinetic  energy 
referred  to  moving,  228. 

Frequency,  of  a  vibration,  263. 

Friction,  46  ;  coefficient  of,  47.  Reac- 
tion between  moving  rough  bodies, 
200. 

Generalized  coordinates,  320,  329 ;  im- 
pulse, 345;  momentum,  345. 

Gravitation,  law  of,  279. 

Gravity,  work  performed  against,  153 ; 
motion  of  body  falling  under,  189 ; 
variation  with  latitude,  200. 

Gyration,  radii  of,  290. 

Hamilton's  Principle,  323. 
Harmonic  motion,  simple,  201. 
Hooke's  law,  44. 
Horse  power,  146. 


Impact,  238 ;  of  particle  on  fixed  sur- 
face, 241 ;  of  any  two  moving  bodies, 
244 ;  of  two  smooth  spheres,  246. 

Impulse,  233  ;  of  compression,  239 ;  of 
restitution,  240;  generalized,  345. 

Impulsive  forces,  233,  345. 

Inclined  plane,  motion  of  particle  on, 
192. 

Indicator  diagram,  151. 

Inertia,  moment  of,  290 ;  coefficients 
and  products  of,  301,  302  ;  ellipsoid 
of,  302 ;  principal  axes  of,  303. 

Inverse  square,  law  of,  276. 

Kepler's  laws,  279. 

Kinetic  energy,  1()8  ;  of  system  of  par- 
ticles, 228 ;  of  rotation,  289 ;  of  a 
rigid  body,  290. 

Lagrange's  equations,  329 ;  for  impul- 
sive forces,  344 ;  for  non-conserva- 
tive systems,  339. 

Lamina,  center  of  gravity  of,  121,  135. 

Latitude,  variation  of  gravity  with, 
200 ;  variation  of  terrestrial,  310. 

Laws,  of  nature,  1 ;  of  motion,  26. 

Least  action,  327. 

Line  of  action,  of  a  force,  60. 

Mass,  and  measurement  of  mass,  29. 

Measurement,  of  velocity,  6  ;  of  accel- 
eration, 12  ;  of  ma.ss,  29 ;  of  force,  30  ; 
of  work,  145  ;  of  acceleration  due  to 
gravity,  195 ;  of  an  impulse,  235. 

Modulus  of  elasticity,  of  a  string,  45. 

Moment,  of  a  force,  00 ;  of  a  velocity, 
274 ;  of  inertia,  290  ;  of  momentum, 
295. 

Moment  of  greatest  compression,  238. 

Moments,  principal,  of  inertia,  301. 

Momentum,  29;  con.servation  of  linear, 
223  ;  moment  of,  295 ;  conservation 
of  angular,  297  ;  generalized,  345. 

Motion,  referred  to  frame  of  reference, 
3 ;  of  a  rigid  body,  91,  280  ;  referred 
to  moving  frame  'of  reference,  197 ; 


INDEX 


363 


of  system  of  particles,  220;  of  cen- 
ter of  gravity  of  any  system,  22-4; 
simple  harmonic,  261 ;  of  particle 
about  a  center  of  force,  269 ;  of  par- 
ticle under  law  of  inverse  square, 
276. 

Neutral  equilibrium,  182. 
Newton's  law,  of  elasticity,  245. 
Newton's  laws,  of  motion,  26. 
Normal  coordinates,  350. 

Orbit,  general  theoiy,  273  ;  differential 
equation  of,  275. 

Orbit  of  a  particle,  law  of  direct  dis- 
tance, 269 ;  law  of  inverse  square  of 
distance,  276. 

Oscillations,  of  a  pendulum,  298 ; 
small,  of  a  general  dynamical  sys- 
tem, 348;  forced,  353. 

Parallel  forces,  96,  99. 

Parallelogram  law,  velocities,  9;  ac- 
celerations, 13  ;  forces,  38  ;  couples, 
105  ;  angular  velocity,  286. 

Pendulum,  simple,  259  ;  seconds,  261 ; 
cycloidal,  265 ;  general  motion  of, 
298. 

Period,  of  vibration,  261 ;  of  simple 
harmonic  motion,  265. 

Plane,  composition  of  forces  in  one,  95  ; 
orbit  about  a  center  of  force  confined 
to  one,  273. 

Planet,  rotation  of  a,  309. 

Point  of  application,  of  a  force,  60, 
95. 

Potential  energy,  163. 

Principal  axes  of  inertia,  303. 

Principal  coordinates,  350. 

Principle,  of  least  action,  327 ;  Ham- 
ilton's, 323. 

Products  of  inertia,  302. 

Projectiles,  205 ;  range  on  a  horizontal 
plane,  209  ;  range  on  an  inclined 
plane,  209 ;  envelope  of  paths  with 
given  initial  velocity,  211. 


Pulleys,  systems  of,  157. 
Pyramid,  center  of  gravity  of,  132. 

Quickest  descent,  line  of,  193. 

Radius  of  gyration,  290. 

Range,  of  a  projectile,  209. 

Reaction,  31 ;  frictional,  between 
bodies  at  rest,  46;  frictional,  be- 
tween bodies  in  motion,  200. 

Reference,  frame  of,  3,  33 ;  motion  of 
frame  of,  197. 

Relative  motion,  4. 

Resonance,  principle  of,  354. 

Rest,  3. 

Restitution,  impulse  of,  240. 

Retardation,  12. 

Rigidity,  90. 

Rotation,  axis  of,  92 ;  of  earth,  198 ; 
of  a  rigid  body,  kinetic  energy  of, 
289  ;  of  a  planet,  309. 

Sag,  of  a  string,  85. 

Sector,  of  a  circle,  center  of  gravity  of, 
129 ;  of  a  sphere,  center  of  gravity 
of,  134. 

Segment,  of  a  circle,  center  of  gravity 
of,  128. 

Simple  harmonic  motion,  2(il. 

Spherical  cap,  center  of  gravity  of,  130. 

Spinning  top,  motion  of,  310. 

Stability  and  instability  of  equilib- 
rium, 174,  351. 

Stars,  double,  orbits  of,  280. 

Strings,  tension  of,  42  ;  flexibility  of, 
43 ;  extensibility  of,  44  ;  on  surface, 
74  ;  sag  of  a  stretched,  85  ;  work  of 
stretching,  150. 

Suspension  bridge,  78. 

System,  of  pulleys,  157  ;  conseiTative, 
of  forces.  163. 

System  of  particles,  statics  of,  59 ;  mo- 
tion of,  220  ;  kinetic  energy  of,  230. 

Tension,  of  a  string,  42,  74. 
Top,  motion  of,  310. 


364 


INDEX 


Transmissibility  of  force,  04. 
Triangle,  of  velocities,  10. 
Triangular  lamina,  center  of  gravity 
of,  121. 

Uniformity,  of  nature,  1. 
Unit,  of  velocity,  6 ;  of  force,  30 ;  of 
work,  145. 

Variation,  of  value  of  gr,  200 ;  of  ter- 
restrial latitude,  310. 

Vectors,  1(> ;  in  one  plane,  16 ;  in  space, 
lit. 

Velocity,  uniform  and  variable,  6; 
average,  0 ;  composition  of ,  7  ;  mo- 
ment of,  274 ;  angular,  286. 


Vibrations.  348,  353. 

Virtual  work,  principle  of,  155. 

Weight,  of  a  particle,  42  ;  of  a  system 
of  particles,  118. 

Wheel  and  axle,  65. 

Work,  measurement  of,  145 ;  against  a 
variable  force,  148  ;  of  stretching  a 
string,  148 ;  represented  by  an  area, 
150  ;  against  an  oblique  force,  152  ; 
performed  against  gravity,  153  ;  per- 
formed by  a  couple,  154 ;  principle 
of  virtual,  155 ;  performed  by  an 
impulse,  235. 

Wrench,  107. 


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ANNOUNCEMENTS 


ELEMENTS  OF  THE 

DIFFERENTIAL  AND  INTEGRAL 
CALCULUS 

With   Examples   and   Applications 
REVISED  EDITION,  ENLARGED  AND  ENTIRELY  REWRITTEN 

By  JAMES  M.   TAYLOR 

Professor  of  Mathematics  in  Colgate  University 


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solutions. 

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